UNIT 8 TAYLOR9$ THEOREM
Structure 8.1 Introduction 5 Objectives 8.2 Taylor's The~rem 5 Taylor's Theorem for Functions of One Variable Taylor's Theorem for Functions of Two Variables 8.3 Maxima and Minima 15 . LocalExtrema Second Derivative Test for Local Extrema 8.4 Lagrange's Multipliers 8.5 Summary 8.6 Solutions and Answers
8.1 INTRODUCTION
In this unit we state, without proof, Taylor's Theorem (about approximating a function by polynomials) for real-valued functions of several variables. This theorem is the principal tool for finding out the points of relative maxima and minima for these functions. We also discuss briefly Lagrange's method of multipliers, which enables us to locate the stationary points when the variables are not free but are subject to some additional conditions. In this unit we will be dealing with functions of two variables. Even though the results are true for any number of variables, their proof involves techniques which are not easy to understand at this level. So, for the sake of simplicity, we confine our attention to the two- variable case. . We start our discussion with the one variable case. Objectives After studying this unit, you should be able to
find the Taylor polynomials for functions of one or two variables, ., state and apply Taylor's theorem for functions of one and two variables, I I locate the stationary points of functions, use the second derivative test to find the nature of stationary points, use the technique of Lagrange's multipliers in locating the stationary points of functions of two variables. ,
8.2 TAYLOR'S THEOREM
In the calculus course you have seen (Unit 6) that if we know the values of a function of one variable and its derivatives at 0, then we can find an expression for the value of the function at a nearby point. We can derive a similar expression for functions of two variables using partial derivatives. This expression was first derived by Brook Taylor, an English mathematician of the eighteenth century. We shall first discuss Taylor's theorem for functions of a single variable. Taylor (1685-1731) 8.2.1 Taylor's Theorem for Functions of One Variable You will agree when we say that polynomials are by far the simplest functions in calculus. We can evaluate the value of a polynomial at a point by using the four basic operations of addition, multiplication, subtraction and division. However, the situation in the case of functions like ex- lnx, sinx, etc., is not so simple. These functions occur so frequently in all branches of mathematics, that approximate values of these fun&ns have been tabulated I ~pplicationsof Partial ' extensively. The main tool for this purpose has been to find polynomials which approximate Derivatives these functions in a neighbourhood of the point under consideration. You are already familiar with Lagrange's mean value theorem. This theorem states that if f(x) is differentiable in some neighbourhood N of the point x,, then we have
for all x such that [x,;x] or [x, x,] is contained in N. Here 5 is a point lying between x, and X.
Iff is twice differentiable in N, then, again applying mean value theorem to the function f, we can go a step further and write
- 1 f(x) = f(x,) + (x-x,) f (x,)+ 5 f' (6) (x-xo)l , where 6 is some point in N lying between x, and x. Thus, the constant polynomial f(x,) approximates f(x) in N in the first case, while the polynomial f(x,) + (x-x,) f (x,)approximates f(x) in N in the second case. The difference between the actual value and the approximated value is called the error term. 1 The error term in the first case is f (5) (x-x,), and in the second case it is - f" (6) (x-xJ2. We 2 can estimate these error terms iff and f" are bounded. Taylor's theorem tells us that if a function f(x) has derivatives of all orders upto n+l in a neighbourhood of x,, then we can find polynomials P,(x), ...... Pn (x) of degree 0, ...... n, respectively, such that the error term f(x) - P,(x) is a polynomial of degree less than or equal to r+l. Note that here we consider the polynomial 0 also as a polynomial of degree zero, which is not the usual practice. We have done this for the sake of uniformity of expression. In order to state the precise result, we 1 start with the following definition. I
' Definition 1 : Let f(x) be a real-valued function having derivatives upto order n 2 1 at the point x, . A polynomial P(x) is said to be the rth Taylor polynomial of f(x) at x,, if
i) the degree of P(x) lr, r I n
ii) P(j)(x,) = f(j) (q)for 0 lj lr, where Po)(x,) = P(x,) and fro, (x,) = f(x,). Recall that a polynomial P(x) is an expression that can be written as
where c,, c,, ...... cn are real numbers. Apart from these there are expressions like P(x) = c, + c, (x-x,) + ...... + c, (x-x,)" . ... (2) where x,, c,, c,, .... are real numbers and x, # 0, which are also called polynomials. You can easily see that (2) can be rewritten in the form (1) by expanding the powers (x-x,)~, .... (x-x,)". We also call the expression in (1) a polynomial it zero and.that in (2), a polynomial at x,.
Now we state and prove a theorem which tells us that Taylor plynomials of a given function are unique. It also tells us how to find out the Taylor polynomials of a given function.
Theorem 1 : Let %. .... a, be any r + 1 real numbers. Then there exists a unique polynomial P(x) such that
i) The degree of P(x) Ir ii) Pc~)(x,)=a~,OSjIr, where x, is any fixed real number. Taylor's Theorem 1 Moreover. P(x) = C,'a, -(x-xJrn. m! I Proof :We can write a polynomial at x, as
where bo ...... , b, are real numbers. Now we have to determine b,, ....., br such that P")(x,) = a, for 0 I jI r. If we differentiate the expression in (3) jtimes, then we get , r Fj)(x) =zk (k- 1) .... (k -j + 1,) 4, (x-x~)~-~,1 I j 5 r, k=j and therefore,
Thus.
Also,
Hence,
PC J ) (x,) "j= j! forOljlr
Substituting for bj s in (3) , we get
Note that the polynomial P(x) will be of degree r if and only if a, # 0. Now by (4) we can conclude that the polynomial is unique. The following corollary of Theorem 1 tells us how to find the Taylor polynomials of a given function. Corollary 1 :If f(x) is a real-valued function having derivatives of all orders upto n (n 2 1)' then the mh Taylor polynomial of f(x) at x, is given by
Proof : Let us take a, = fi) (x,), 0 5 k .5 m, in Theorem 1. Then the mh Taylor polynomial off, if it exists; must be in the form of Equation (5). Thus, \
The above discussion shows that the Taylor polynomials of a given function can be found step by step using. the relation
1 MOrebver, if P,,,(x) is the mb Taylor polynomial of f(x) at x, ,then you can check that the derivative of Pm(x) at x, is the (m-I)* Taylor polynomial of f(x) at x, . Let us consider some examples now. Example 1 : Let us find the Taylor polynomials of
We apply Theorem 1 with x, = 3. . Applications of Partial since eO)(3) = f(3) = 22, Derivatives e') (3) = 19,
en (3) = 6 and
19 Po (x) = 22, P, (x) = 22 + (x - 3),
19 14 6 P3 (x) = 22 + (x-3) + -2! (x-3)' + -3! (x-3)l and
P, (x) = P3 (x) for all r > 3. b Example 2 :Let us find the fourth Taylor polynomial
1 1 We have f(x) =- (1 + x)-' 2
1 1 3 15 Therefore, f(0) = 1, f (0) = , f" (0) = - (0) = - ti4)(0) = - - 2 -4 ' e3) 8 ' 16 - The desired polynomial is
f (0) f"(0) (0) x3 + ff4I (0) 4 T4(x) = f(0) +--- x+- x2+ l! 2! 3! 4!
Example 3 : Let us find T, (x) for cos x at x, = z.
Now cos x = - 1 and the first eight derivatives of cos x at x are
Dropping the terms with coefficients 0, we have the polynomial
1 .Example 4 :Let us find T5 (x) at x, = 0 for f, where f(x) = -= (1 - x)-' 1-x Computing the derivatives, we obtain
f. (x) = (1 - x)-~,f' (x) = 2(1-~)-~.
r3)(x) = 3.2 (I-X)~, f4' (x) = 4 ! ( l-x)"
P" (x) = 5! (1 - x)~ Thus, the successive derivatives off at 0 , in order, are Since f(0) = 1, we obtain Taylor's Theorem
= 1 +x+x2+x3+x4+x5 Now you can try these exercises.
El) Find the nthTaylor polynomial of the function ex at x = 2.
E2) Find the 6" Taylor polynomial of sin x at x = 0.
E3) Find the rh Taylor polynomials of the following functions at the indicated point and for the indicated value of r.
E4) Find a polynomial f(x) of degree 2 that satisfies f(1) = 2, f (1) = -1 and f' (1) = 2.
We now state Taylor's theorem which gives us the connection between a function and its Taylor polynomials at a point. Theorem 2 (Taylor's Theorem) : Let f be a real-valued function defined on the open interval ]a, b[. Suppose f has derivatives of all orders upto and including n + 1 in the interval ]a, b[. Let x, be any point of the interval ]a, b[. Then for any x in ]a, b[,
(x-x0 )" + f(x) = f(~,) +T (x-x, ) f(xo) + .... + wn! p, (%)+ (n+l)! PI) (c), where c is a point between x, and x. The expression on the right hand side of (6) is called Taylor's expansion of f(x) at x,.
We won't give the details of the proof of this theorem here. But we wish to indicate that it can be proved by applying ~olle'stheorem to the function
=f (x)+ ...... + Fn)(X) + (x-X) + A, defined on the 6x)=f(x) + I ! n! interval [x,,x] or [x, x, 1 according as x, < x or x < x, , where A is a constant so determined that @(x,)= @(x). The point c in Theorem 2 comes from the application of Rolle's theorem to the function @ (X) and therefore we can only assert its existence and not the exact location.
Now we rewrite Equation (6) in the form
f(x) = Pn (x) + R .+ 1 (x), where Pn (x) is the nh Taylor polynomial of f(x) at x, and
Then R,,+ ,(x) depends on x, xo and n. We call R, + , (x) the Lagrange's form of remainder after n+l terms in the Taylor's expansion of f(x) at xo . If we write x as %+h, then Taylor's expansion becomes
n p) hr hn+l f(xo + h) = - - + n! f (n+ l) (xo + Oh), m (x,) r ! where 0 < 0 < 1, and 0 is a real number depending on x, and h.
Note than if f(x) is a polynomial of degree m, then fir) (x) = 0 for r > m. Therefore, %+ (x) = 0 for all x and x, ,provided n > m. Thus, in this case, finding Taylor's Applications of Partial expansion off (x) upto m+l terms at x, is equivalent to expressing f(x) as a polynomia1,in Derivatives x-x, with coefficients from R. By estimating the remainder R,,+, (x), we can find how close is f(x) to its nthTaylor polynomial. In Calculus (Unit 6) you have learnt how to write Taylor's or Maclaurin's series of a given function. At that time you were cautioned that these series need not be valid for a given function. In fact, the Taylor series of a function has a very close connection with its finite Taylor's expansion. Let us see. Suppose f(x) has derivatives of all orders at x, . If
is Taylor's expansion off at a point x, , and if we are able to prove that
lim % + , (x) = 0, n+ - then we say that the Taylor's series of f(x) at x, converges to the given function for all those x for which lim %+,(x) = 0. n+ - Moreover. in that case we write - P' (x, ) f(x) = -(X - Xo )". & n!
The coefficient is called n-th Taylor's coefficient in the Taylor's expansion of f(x) at x, . We now illustrate Taylor's theorem with the help of a few examples. Example 5 : Let us apply Taylor's Theorem to the function f(x) = ex about x = 0 in the interval ] - 1, 1 [. Yoh know from Calculus that the function f(x) = ex is continuous everywhere on the real line, and
f(x) = f (x) = ...... = P' (x) = ...... = ex.
Thus, derivatives off of all orders exist and are continuous in the interval ] - 1, 1 [. Then by Taylor's theorem, given any x E ] - 1, 1 [ and n E N, there exists a point c between 0 and x such that
where
Now can we say anything particular about the remainder Rn+ , (x) in the above example ? Let's see.
Now since c lies between 0 and x, we have ec < elx'.
Then I R. + , (x) ( c elx\ l&l
Now -can be made as small as we like by choosing n sufficiently large, (n+l)! 1 i.e., lim -= 0. n + - (n+l)! Taylor's Theorem This means that R,,., (x) + 0 as n + m.
Thus, for the function f(x) = exwe can write
Let us consider another example.
Example 6 : Let us find Taylor's expansion of f(x) = In (1 + x) for x E ] -1, 1 [ at x = 0.
You know from Calculus (Unit 6) that
sothat P)(O)= (-1)"-I (n- l)!
Hence,
(-1)" 1 Xn+l. where R,, + , (x) = n + 1 (1 +C)"+'
Clearly for x E 1 - 1, 1 [,
This shows that lim R,, +,(x) = 0. n + - - 1-1)n-1 xn Thus, In (1 + x) = Z foranyx~1-1, 1[. n=l n !
Try the following exercises now.
1 E5) Obtain the Taylor's expansion of f(x) = about x = 0 in the interval ] - - , 1[. 1+x 2
X E6) Obtain the Taylor's expansion of f(x) = sin x about x = - 6
So far we have seen how to find Taylor's expansions for functions of a single variable. In the next sub-section, we shall discuss Taylor,'s theorem for functions of two variables. 8.2.2 Taylor's Theorem For Functions of Two Variables In this sub-section we extend Taylor's theorem to functions of two variables. For this let us first extend the notion of Taylor polynomials to functions of two variables. You have already seen the definition of a polynomial in n variables in Sec. 3.3. Here we will discuss the polynomials in two variables in detail. Definition 2 : Let x and y denote two variables. Then an expression of the form aJkxJyk, I where j and k are non-negative integers and a,,€ R, is called a monomial. The integer j + k is called the degree of the monomial. For example, xZy3is a monomial of degree 5,
x4 is a monomial of degree 4, I ! y7is a monomial of degree 7, F I A polynomial in x and y is nothing but a finite sum of monomials. We now give a formal definition. Applkations of Partial Definition 3 : A polynomial in two variables in x and y with coefficients in R is an Derlvatives expression of the type P(X,Y)= a, + (alox + %,y) + (w2+ a,,xy + %y2) + ...+ (%,xi + a(,- ,,, x -I y + ..... + a,,,yi) + .. ... + (a,&" + ..... + a,,,yn). where aij's are real numbers.
Here you can note that we have grouped together the monomials having the same degree. In the fust bracket each term is a monomial of degree 1. In the second, each is a monomial of degree 2, and so on. For example, P(x,y) = 1 + 2xy + x2y is a polynomial in two variables. This polynomial is a sum of three monomials, having degree 0.2 and 3, respectively. The number 3, which is the maximum of these numbers is called the degree of this polynomial. In general, we have the following definition. Definition 4 :The highest degree of the monomials present in a polynomial P(x,y) is called the degree of P(x,y). You can now easily do this exercise.
E7) Find the degree of the following polynomials:
a) 1 + y + x2y + xy2 + ys b) 2 + x3 + y3 c) 7 + x + xy t x3y + x4
Now we give the definition of the nh Taylor polynomial of a function of two variables. Definition 5: Let f(~,~)%ea real-valued function of two variables. Assume that it has continuous partial derivatives of all types of orders less than or equal to n in some neighbourhood of a point (x,,, yo). Then
is called the nth Taylor Polynomial off at (x,,, yo). In particular, if f(x,y) is a polynomial of degree n, then all partial derivatives of order m for m > n will be zero. Therefore T, (x.y) = Tn (x.y) for all m 2 n. Further, as in the case of one variable. you can see that Tn (x.y) at (0.0) is equal to f(x,y). - Again. from the definition, you can see that
so that the Taylor polynomials of a given function f(x,y) can be computed step by step. We show this by an example. Example 7 :Let us find the Taylor polynomials of the function P(x, y) = 1 + 2xy + x2y at (1.1). We first note that P(1.1) = 4 and therefore, To (x,y) = P( 1.I) = 4.
x-i ap (y - 1) dP Therefore. T, (x,y) = To (x,y) + (1.1) + y ax l! dy (1. 1) = 4 + 4 (x -1) + 3(y - 1). Taylor's Theorem 2P #P #P =&T =2y , -= 2 +2x, 7 = 0 ax ay ay (x - 1)2 BP x-1) (y-1) BP Therefore, T2 (x,y) = T, (x,y) + 2! )+1 1 dx dy (1, 1) BP
= 4+4(x-1) +3(~-1)+(~-1)~+4(~-1)(y-1). 2P 2p 2p 2p Since - = 0, -= 2, ------0 and-- = 0,wegetthat dX3 dx2dy dxdy2 dy3 T3 (x,Y) = T2 (XJ) + (X- (Y- 1) You will agree that T,(x,y) = T3 (x,y) for all r 2 3. Here is another example. Example 8 : Let us find the Taylor polynomial T, (x,y) for the function sin (x + y) at (0, 0). We write f(x,y) = sin (x + y). It is clear that f has continuous partial derivatives of all orders. Let us compute these derivatives at (0,O). We get df df & (x'Y) = cos (x + y) = - (x, y) dy df X Therefore - (0,O) - - dX -& (O*O)= '. -d2f 2.f Bf &2 . (0,O) = -(0, 0) = - (0,O) = - sin (x + y) = 0 ax dy dy2 (0,O)
Thus, the third Taylor polynomial of sin(x + y) at (0,O) is
When simplified it takes the form
Try these exercises now.
E8) Find the second Taylor polynomial of ex + y at (0.0). E9) Find the Taylor polynomials of f(x,y) = 2 + x3 + y3 at (1,l). E10) Let f(x.y) be a polynomial of degree 2. Prove that T2(x,y) at (0.0) is equal to f(x,y).
Now let us consider a function f(x,y) of two variables. Assume that f has continuous partial derivatives of all orders less than or equal to n, for some integer n, in a neighbourhood of a . point (xo. yo). Then the nth Taylor polynomial . ..>. Applications of Partial Derivatives
has the same value gs f(x,y) at (%,yo), and the same partial derivatives of all orders S n as f at (x,, yo). As in the case of one variable, we would naturally like to know whether we can approximate f by the corresponding Taylor polynomials. Put differently, we would like to have some information about the function
An analogue of Taylor's theorem whicd we state now, provides us some information about the function R,,+ , (x, y). Theorem 4: (Taylor's Theorem For Function of Two Variables): Let f be a real-valued function of two variables x and y with continuous partial derivatives of orders In + 1 in some neighbourhood S (x ,r) of x = (x, ,yo). Then for a given (x,y) # (x, ,yo) in S (x ,r) , there exists a point (cI,c2)on the line segment joining (%,yo) and (x ,y), such that
where
and
This means %+,(x ,y) = .(x-xg )"+I
So you can see that %+ (x, y) involves all the (n + 1)'"order partial derivatives off evaluated at the point (c, ,cz).
The right hand side of (7) is called the nth Taylor expansion off at (x, , yo). This expansion may seem a little complicated to you. But don't let it scare you. You will soon see that in this course you need to consider only the second Taylor expansions of functions. If you look at the expression for R, (x,y), you will see that it contains powers of (x-%) and (y - yo). Now if we take the point (x ,y) close enough to (x, ,yo), then (x - xg) and (y - yo) will be very small. Therefore, we can get a good enough approximation of f(x,y) by a second degree polynomial. Of course, f(x,y) can be approximated as closely-as we like by a polynomial by choosing n sufficiently large.
For future use, we write the expression for T2(x,y) and the second Taylor expansion of f(x,y) at (%,yo)explicitly : Taylor's Theorem
Now consider this example. Example 9 :Suppose we want to find the second Taylor expansion of the function f(x,y) = In (1 + x + 2y) for points close to (2,l).
Let us compute the partial derivatives one by one. We have f(2, 1) = In 5
df 1 df 1 - - (271) =- - 1 + + 2y 9 5
df 2 df 2 - - - (2,l) = - dy- 1+x+2y 'dy 5
2f -1 2f -1 -- (27 1) = - &2 - (1 + + Zy)2 * 25 2f -4 2f -4 -- - (2, 1) = - dy2 - (1 + x + 2~)~' dy2 25 2f 2 2f -2 -- (2, 1) = - dxdy - - (1 + x + 2~)~' dxdy 25 Then the second Taylor expansion is given by
2 4 ' 2 ( - T (x -xo) (Y - YO) + (- z)(Y - Yo)2]
Why don't you try some exercises now?
El 1) Find the second Taylor expansion for the function f(x, y) = xy2 + cos xy IC about (1, - ). 2
E12) Find an approximation to the function f(x,y) = exsiny by a second-degree polynomial near (0,O). (I
In the next section we will discuss maxima and minima for functions of two variables. You have already studied maxima and minima for functions of one variable in Block 2 of your Calculus course. There you have used the first and second derivative tests to locate the local maxima and local minima. You will see that the definitions of the maximum and minimum of functions of two variables are similar to those in the case of one variable. We shall obtain a set ::f necessary conditions for the existence of maxima and minima, which are similar to the one variable case. Taylor's expansion of functions which you studied in this section will enable us in deriving a set of sufficient conditions for determining the points of maxima and minima. This is an analogue of the second derivative test, which you have studied for functions of open variable.
83 MAXIMA AND MINIMA
This section deals with the concept of maxima and minima for functions of two variables. You know from the one-variable case that the study of maxima and minima (or extrema) is useful in graphing a function.
As in the one-variable case, we shall be interested in studying the local extrema of a function, rather than its absolute extrema. So let us take up the study of local or relative extrema, i.e., points of relative maxima or relative minima for functions of two variables. Applications of Partial 8.3.1 Local Extrema Derivatives We shall first try to understand the concepts of local maximum and minimum for functions of tbo'frariables. Let us start with some simple functions. Consider the function f(x,y) = (x + 1)Z+ (y - 3)Z - 1 NOWf(-1.3) = -1.
Since (x + 1)' and (y - 3)' are always positive for x # -1, and y # 3,
we have (x + 1)' + (y - 3)' - 1 > - 1 for (x, y) # (-1, 3). That is, f(x, y) 2 f(-1, 3) for all (x, y). In this case we say that f has a minimum at (-1, 3). See Fig. 1.
Fig. 1
You can see that the tangent plane to the surface at (-1,3, -1) is horizontal. Now consider another function defined by 1 f(x, y) = 5 - sin (xZ+ yZ)
1 X; Here f (0,O) = 5 . Take the circle xZ+ yZ= - with centre (0,O). Then for any (x, y) (0,O) 6 * inside the circle, we have
sin (xZ+ yZ) > 0 and therefore 1 1 f(x, y) = - - sin (xZ+yZ) < - = f(0,O). 2 2
Thus f(x,y) I f(0,O) for all (x, y) in the circle. Note that f(x,y) can be greater than for 2 (x, y) outside the circle. In this case we say that f has a local maximum at (0,O). This leads us to the following definition: Definition 6: Suppose f is a real-valued function of two variables. We say that the function f has a local maximum at a point P(x,,y,) if there exists some open disc S( x , r), wnere x = (x, ,yo) and r > 0, contained in the domain of definition off such that Taylor's Theorem for all (x, y) E S (X ,r) we have
f(x, Y) 5 f(x, 9 yo).
Now we are sure you will be able to define a local minimum on your own. See E 13). Don't forget to tally your definition with the one given in Sec. 8.6.
------E13) Define a local minimum of a function of two variables.
Recall that in the case of functions of one variable, we took an open interval instead of an open disc. Thus, our notion of local maximum and local minimum for functions of two variables is a natural generalisation of the notion in the one-variable case.
Remark 1: i) If f(x,y) 5 f(xo ,yo) for all (x, y) E D, the domain off, then we say that the point (xo,yo)is the global or absolute maximum of f. We can similarly define a global minimum. ii) The maximum and minimum values of a function are called extrema of the function; we say that a function has an extremum at a given point if the function has a maximum or a minimum at that point. If a function has an absolute maximum or minimum at a point (xo,y,) and this point is such that there is a neighbourhood of (%,yo) contained in the domain off, then (xo,yo)is a relative, maximum or relative minimum of the function f. But the converse may not be true. In the case of functions of a single variable you know that the derivative of a function vanishes (if it exists) at each local maximum and minimum. There is a similar result for functions of two variables also. We present this result in the next theorem.
Theorem 5: Let f be a function of two variables. Suppose f has an extremum at some point (%,yo) and the partial derivatives off exist at that point. Then
-df af (%. Yo) = 0 = -ay (x, ,yo). Proof : Let us assume that f(x,y) has a relative maximum at x = (x,,yo). Then f(x,y) is defined in an open disc S = S (x, r), r > 0 and f(x,y) If(%, yo) for all (x,y) E S. Thus, there exist open intervals I, and I,,
such that x E 11 + (x,yo) E S and y E I, * (xo,y) E S. Now consider the function g, defined on I, by
Then g, is a function of one variable. Similarly, the function g, defined on I, by g, (y) = f(xo,y)8. 1s also a function of one variable. We can see from the definitions of g, and g, that
gl (x) = f(x,yo) 5 f(xo,yo)= g,(xo) for all x E I, ,and
g2 (Y) = f(x,,y) If(xo,yo) = g, (yo) for all y E I,. This means that the functions g, and g, have relative maxima at the points x, and yo, respectively. Now we are given that the partial derivatives off exist at (xo , yo). This means that g, and g, are differentiable at xo and yo, respectively. Thus,
g,' (xO)= f, (x0 , YO ) = 0 and because you already know that if a one-variable function has relative extremum at a point and is differentiable at that point, then its derivative vanishes at that point. Applications of Partial If f(x,y) has a relative minimum, then g, and g2 also have relative minima at x, and yo, Derivatives respectively, and we have the same conclusion as above. We can use this theorem to check whether a given function has an extremum at some point or not. All we have tr. .: 7 is to see whether its partial derivatives vanish at that point (if they exist). We shall illust: ,.&this fact with some examples. Example 10 : Let us check whether the function given by
has maximum or minimum values.
The iiven function f(x, y) = x2 - 2x 'is differentiable everywhere. According to + 4 Theorem 5, first we have to find out the points (x, y) such that f, (x, y) = 0 = fy(x, y).
Now
Therefore, f, (x, y) and fy (x, y) will vanish only when x = 1 and y = 0. Therefore, the point (1, 0) is the only possible point where f can have a maximum or minimum value. Now let us see whether (1,O) is a maximum or a minimum point for f. We rewrite f(x, y) as
This shows that, f(x, y) 2 -1 = f(1,O) for all (x, y). Thus, the function has a global minimum at (1,O). The minimum value is f(1,O) = -1. The function has no maximum values. df df Thus, if - # 0 or -# 0 at some point, then we can straightaway say that the function dx dy df df does not have an extremum at that point. But if - = 0 = - at some point, then this dx dy does not imply that the function has an extremum at that point. It is possible that all the first order partial derivatives of a function are zero at some point (x, , yo), but still, that point is not an extremum point for that function. That is, the converse of Theorem 5 is not true. We illustrate this with an example.
Example 11 : Consider the function f : R2 + R defined by
For this function we have df df - - -2x and - - 2y. dx - dy - df df Therefore, -ax (0,O) = 0 = -ay (0,O). Now, let us check whether f has an extremum at (0,O). We have f (0,O) = 1, f(x, , 0) < 1 and f(0, y,) > 1 for all non-zero x, and y,. But in any neighbourhood of (0,0), we can find points of the type (x, ,0) and (0, y,). Thus, there exists no neighbourhood N of (0,O) for which f(x, y) If(0,O) or f(x, y) 2 f (0,O) for all (x, y) E N. (also see Fig. 2).
Thus, (0,O) is neither a maximum nor a minimum point for f, though both the partial derivatives off vanish at (0,O). In Fig. 2 you can see the graph off. Taylor's Theorem
Fig. 2
Sometimes it may happen that the partial derivatives of a function do not exist at a point, but, still the function has an extremum at that point. This is the case with the function in our next example. 't Example 12: Consider the function given by
f(x, y) = 1 + In Fig. 3 you can see the graph of this function. Obviously, f has a minimum at (0,O). df df Now let us try to calculate -dx (0,O) and -ay (0.0). 1 df f(h,O) - f(0,O) - (0, 0) = lim dx h +O h 1 47-1 = lim + h -to h lhl = lim - . h+O h Ihl df Fig. 3 But we know that lim - does not exist. Hence - (0,O) does not exist. You'can h+O h dx df similarly check that -ay (0,O) also does not exist. See if you can solve these exercises now.
E14) Find the points at which the partial derivatives of the function X f(x9~)=X2+ y2 - 4 vanish.
E15) show that the function f(x, y) = x2 + y2 - 8x - 2y + 18 has a global minimum. (Hint : Complete the squares involving x and y.)
Uptil now we have seen that if a function f has an extremum at (a, b), then either i) the partial derivatives off do not exist at (a, b), or df (a, b) = Q = dt (a. b). ii) ay Henceforth in this unit we will be concerned only with functions whose partial derivatives exist. So the second condition becomes a necessary condition for the existence of extrema. Because of the importance of this condition, we give a special name to the points satisfying it.
Definition 7 : Let f be a function of two variables. A point (x, y) is said to be a stationary point off if both the partial derivatives are zero at (x, y). Now you will agree that iff has an extremum at (a, b), then (a, b) is a stationary point off. But all stationary points of a function need not be its points of extrema. You have seen such a situation in Example 11. You can now try this exercise.
- - E16) Find the stationary points of the following functions: a) f(x, y) = 1 + x2- y3
b) f(x, y) = (x + y) e -xY
Now let us look at some stationary points which are not points of extrema. Suppose (a, b) is a stationary point of a function f(x, y) but is not its point of extremum. Still it is possible that one of the functions f(x, b) or f(a, y), where a and b are fixed can have a maximum at (a, b), while the other has a minimat this point. For example, let us look at the surface given in Fig. 4 by the equation
(0,O) is a stationary point of f(x,y). Look at the surface along x and y axes. If we fix y = 0, then f(x, 0) = -x2 considered as a function of one variable, has a maximum at 0. But f does not have a maximum at (0,O). And if we fix x = 0 ,then f(0, y) = y2 has a minimum at 0, but f does not have a minimum at (0, 0). Such stationary points are termed a3 saddle points. From Fig. 4, you can see that the graph of f resembles a saddle around the stationary point (0,O). That is why f does not have an extremum at (0. O), even though f, (0,O) = fy (0,O) = 0. We say that a function f has a saddle point at (x, ,yo), if there is a disc centred at (x, ,yo ) such that i) f assumes its maximum value on one diameter of the disc only at (x,.yo). ii) f assumes its minimum value at another diameter of the disc only at (%,yo). Thus, we have seen that a stationary point may not be a point of extremum. In the next sub- section we try to find conditions under which a stationary point is a point of either maximum or minimum. 8.3.2 Second Derivative Test For Local Extrema In this sub-section, we derive a method using which we can test whether a point is a point of maximum or minimum. You will see that this test involves second derivatives. You may recall that for the case of one variable also, we have a second derivative test for testing maxima and minima. According to this test, if we have a function f of one variable such that f (x,) = 0, then f has
a local minimum at x, iff' (x,) > 0
a local maximum at x, iff' (h)< 0. We have a similar test for two variables. But the test is not as easy as in the case of one variable. You are already familiar with homogeneous functions (see Unit 7). In order to find a set of Teglott Theorem sufficient conditions for determining the nature of stationary points, we shall need a simple result about .the sign of values assumed by homogeneous polynomials in two variables of degree two. We shall call a homogeneous polynomial of degree two in n variables with real coefficients, a real quadratic form in n variables. A quadratic form in two variables is also called a binary form or a binary quadratic form. Thus, a binary form is an expression of the type ax2 + bxy + cy2, where a, b, c are real numbers. Now we state and prove a theorem which says that we can determine the sign of a quadratic form by looking at its coefficients. Theorem 6 : Let q(x, y) be a binary quadratic form. Then
ii) b2 - 4ac > 0 * q (x, y) takes positive as well as negative values. iii) b2-4acc Oanda>Oorc>O*q(a,p)>O tta,p~,R(a,f3)#(0,0). iv) b2-4acc Oanda Proof : i) if b2 - 4ac = 0, then both a and c cannot be zero. Let us assume without loss of generality that a # 0. Then This means that q(a, p ) has the sign of a, if it is non-zero. In fact, we obtain that q (a, p) 2 0 for all a, p E R if a > 0 and q (a, p) r 0 for all a , p E R if a < 0. (No-te that in this case there exist a , p E R such that (a ,p) # (0.0) b and q(a, p) = 0. For example, a = - - and = 1.) 2a P By interchanging the rol6f x and y, we can prove that q(a,f3)2Oforalla,p~Rifc>O and q (a. fJ)I 0 for all a,p E R if c c 0. ii) b2 - 4ac > 0. If both a and c are zero, then q(x, y) = bxy, and therefore q(1, -1) = -b, q(-1, -1) = b. This sh6ws that q(x, y) assumes both positive and negative values. \ Now suppose a # EThen Also, if a, = 1 and P, = 0, then q(al fJ1)= a. Thus, q (a, p) and q(al, p,) have opposite signs. This proves Case ii) if a # 0. Similarly we can prove Case ii) when c # 0. (iii) and (iv) : If W - 4ac< 0, then neither a nor c can be zero. Since a > 0 <=> c > 0, it is enough to prove the result in case a > 0 or a < 0: As before, we write I Applications Partial I Derivatives Then for a, i3 E R,the expression inside the bracket is positive for (a, P ) + (0.0) and is zero for (a,p) = (0,O). Thus, if a > 0, q (a, P) 2 0 for every a. i3 E R and if a < 0, q (a, $) 5 0 for every a, p E R. This completes the proof of Theorem 6. Now we use this theorem to obtain a sufficient condition to determine the nature of stationary points. Theorem 7 : Suppose f (x, y) is a function of two variables such that f (x, y) has continuous partial derivativqs upto order two in a disc N containing the point (x, ,yo). Suppose that the point (x, , yo) is a stationary point off (x, y), that is, d2f 2dz f dtf Further, let -(xo , yo ) = a, -(%,yo)=b and-. (x,.Yo)=c. ax2 may ai2 Then, i) f(x,y) has a minimum (maximum) at (x, ,yo ) if bZ- 4 ac < 0 and a > 0 ii) f (x, y) has neither maximum nor minimum at (x, .yo) if bZ- 4ac > O,.i.e., (x, ,yo) is a saddle point. Proof : Let us first consider the functions These functions are given to be continuous at (Q ,yo) E N, and g, (% ,yo) = a. 2gZ(q.yo)= b, g3(x, .yo) = C. This implies that the function g:-g,g, is also continuous on N, and Therefore, by Theorem 6 of Unit 4, there exists a neighbourhood N, of (x,, yo) contained in N such that the function g; -gl g3 will have the same sign as (gi - g,g3) (x, ,yo) on N1. Similarly, (i) Corresponding to g, we can choose a neighbourhood Nzof (x,, yo) contained in N such that the function g, will have the sane sign as g1(x, , yo) in Nz. (ii) Corresponding to gz we can choose a neighbourhood N, of (x,,yo) contained in N such that the function g, will have the same sign as g, (x, ,yo) in N,. Let No = N, n Nzn N, . Then f satisfies all the hypotheses of Taylor's theorem in No. Therefore by second Taylor expansion we have df + (xo,YO) (YO+ k- Yo) dy 1 Taylor's Theorem where h and k are numbers such that (x, + h ,yo + k) belongs to No and (5, 'tl) is a point on .ne line segment joining (x, ,yo) and (xo+ h, yo+ k). Since (xo ,yo ) is a stationary point we have Therefore, 1 f(x0+ h, yo + k) - f(x0 yo ) = 2 [ g12 (5 q) h2 + 292 (6- q) + g: (69 q) k2I = q (h, k) , say . .. (8) Then q (h, k) is a quadratic form in h and k. Now let us take up the cases (i) and (ii), one by one. 2 b2 - 4ac Case (i) : Suppose W - 4ac < 0 and a > 0. Then (g2 - g, g3) (x0 , yo) = < 0. Thus, gi- g, g, will be negative at all points in No, in particular, at (5, q) E N,. That is, Also, since a = gl (x, ,yo)> 0, using the same argument we can conclude that gl(5.N > 0. Therefore, by Theorem 6, the quadratic expression q (h, k) 2 0. This shows that f(x0 + h, yo+ k) 2 f(xo, YO) for all h, k such that (x, + h, yo + k) E No. Hence f has a local minimum at (x,,yo). Similarly, we can show that if c > 0, then f has a local minimum at (x,,yo). Using a similar argument we can prove that f has a local maximum if a < 0 or c < 0. Case ii) Suppose bz- 4ac > 0. b2- 4ac = 4 (gi- g, g3 ) (x,. yo) > 0 implies that g:-gl g, is positive at all points in No. Therefore, by condition (iii) of Theorem 6, q (h, k) takes both positive and negative values. Hence from Equation (8). we conclude that f does not have an extremum in this case. This proves Case (ii) of the theorem. Here, you may wonder why we don't consider the case b2 - 4ac = 0. If b2 - 4ac = 0, then we cannot conclude that g;-g, g, = 0. In fact, the quadratic form could have different signs at different points of No. Thus, the case b2- 4ac = 0 is a doubtful case. Now we are giving a few examples to show how Theorem 7 can be applied to determine the points of extrema. Example 13 : Let us find the.loca1extrema of the function f(x, y), where f(x,y) = x2- 2xy + 2yZ- 2x + 2y + d. Here we have Applications of Partial For both the partial derivatives to be zero, we must have Derivatives 2x-2y-2=0 and -2x + 4y + 2 = 0. Adding these equations, we find that 2y = 0 or y = 0. Then we must have x = 1. Thus, (1, 0) is the only stationary point. Computing the second derivative, we get Therefore, b2 - 4ac = -16 c 0. Since a = 2 > 0, by Theorem 7, f has a local minimum at (1,O). Example 14 : We shall show that the function., f(~,Y) = x2 - ~XY+ y2 + ~3 - ~3 + 2~7 has neither maximum nor minimum at (0, 0). Clearly fx (0,O) = 0 = fy (0, 0) . i.e., (0, 0) is a stationary point. 2f 22f Moreover, a = 3(0.0) = 2. c = -7 (0, 0) = 2, b = -- (0, 0) = - 4, so that aY axay b - b2- 4 ac = 0, showing that Theorem 7 is not applicable. But f(x, y) = (x - Y)~+ (x - y) (x2 + xy + yZ)+ 2x7 Therefore. Consequently in every neighbourhood of (0, O), there exist points (x, 'y) with y = x such that f(x, ~)~ss"mesboth positive and negative values, showing that f(x, y) has neither maxisum r.or minimum at (0,O). Now we give an example of a function, where b2- 4ac = 0 at (xo,yo), but the function has an extremum at (xo, yo). Example 15: Consider the function f(x, y) = y2+ x2y + x4 df - df Here -ax (0, 0) = 0 = -a~ (0, 0). Therefore, 4 (%)l- 4 (-$1 (&)JY = 0 at (0.0). x2 3 But f(x, y) = ( y +- )' + - x4. Therefore, 2 4 -f(x, y) 2 0 = f(0, 0) for all x, y, showing that (0, 0) is the minimum of f (x, y). Now we shall see an application oflhe concept of maxima and minima. Example 16: Of all the triangles of a fixed perimeter, let us find the one with maximum area. ;If the sides are x, y, z, then the area A is given by the formula .. A'= s (s-X) (s-y) (s-z), 1 where s = - (x + y + z) is the semi-perimeter. 2 Taylor's Theorem Here s is a constant and x and y are variables. Therefore, in order to maximize A, it is sufficient to maximize f (x, y) = (S - X) (s - y) ( x + y - s). Now, h Since in a triangle, s + x 'hd S * y, df df the equations - = 0 = - imply that dx dy - Consequently, 2 x=y=- S. 3 This gives the stationary points off. I For these values of x, y, we get b= 2-=--- CYf 2 s dxdy . 3' 2s These conditions ensure a maximum. Now when x = y = - , we get 3 Hence x = y = z, i.e., the triangle is equilateral. b In the next section you will study a method for finding the maximum or minimum of functions of several variables subject to certain constraints. But first it is time to solve some exercises. E17) Find the stationary points and the local extreme values for the following functions: a) f(x,y)=x2+2y2-x. b) f(x, y) = x2+ y3+ 3xyZ- 2x c) f(x, y) = y + x sin y Applications of Partial E18) Discuss the behaviour of the function Derivatives f(x, y) = 2 cos (X+ y) + exY at the origin. E19) Let n be an integer, n 1 2, and let f(x, y) = axn+cy", where ac # 0. a) Find the stationary points off. b) Find the local extreme values, given that (i) a > 0, c > 0, (ii) a < 0, c < 0. 8.4 LAGRANGE'S MULTIPLIERS Supposk we want to construct a closed box in the form of a parallelepiped of maximum volume using a piece of tin of area A. Let x, y, z denote the length, width and height of the box, respectively. Then the problem reduces to finding the maximum of the function f (x, y, z) = x y Z, given that 2 xy + 2 xz + 2yz = A .. . (9) In Fig. 5, you can see a closed box. xyz is the volume of this box and 2xy + 2xz + 2yz is its surface area. , In this section we shall study such problems for functions of two variables, where the variables satisfy some side conditions as in (9). That is, we will discuss a method to find out the maximum and minimum values of a function Fig. 5 z = f(x, y), given that x and y are connected by an equation If we could eliminate one variable from the equation z = f(x, y) with the help of the relation g(x, y) = 0, then z would become a functinn of one variable. Then we can easily find its extreme values. 1, So, the problem reduces to finding the maximum and minimum values for a function of one variable. Here is an example to illustrate this. Example 17: Suppose we want to find the extreme values of the function f(x, y) = x2 + 2y2 -X on the unit circle x2+ y2= 1. We first use the constraint x2 + y2= 1 to reduce the function f (x, y)= x2 + 2yZ-x to a one-variable function. Thus, we get Here we have a one-variable function, say g(x) = 2 - x2-x, defined on the interval (-1.1 1. Now we shall find out the points of extrema for g(x). By solving g' (x) = -2x - 1 = 0, we 1 1 get that x = - is a stationary point of g(x). Then to check whether x = - - is a maxinium 5 2 or a minimum, we calculate 1 Taylor's Theorem Thus, g" (- -) c 0. Therefore, by the second derivative test for one variable we get that 2 1 x = - - is a point of maximum for g. g has no minimum. Now we substitute the value 2 x = - in (10). Then we have '2 1 43 Therefore we conclude that the function has a maximum at two points ( - 2 , + -y) and 9 Thus, the maximum value of the function on the unit circle is - 4 . You must have found this example quite easy to follow. But it is not always feasible to use this procedure. The reduction of the given function to a function of one variable using the given constraint might prove to be quite cumbersome or sometimes might not be possible at all. We now present an alternative method which is often more convenient. This method is known as the method of Lagrange's multipliers. Suppose we want to maximize or minimize a function z = f(x, y) subject to the condition g(x, y) = 0. Theoretically, z is a function of a single variable (say x) and at the extreme dz values - = 0, i.e., & From the relation g(x, y) = 0, we find that at the extrema, we must have Multiplying Equation (12) by an undetermined multiplier h and adding this to Equation (1 I), we obtain Choose h so that the coefficient of vanishes in (1 3). (This is possible and would become clear after you have studied the implicit function theorem in Unit 10.) Hence at the points of extrema, we have From these equations, we can determine the three unknowns x, y, h . The values of x, y give us the coordinates of the stationary points. The role of h is over and we don't need it any i more. We may add here that each stationary point so determined need not be a maximum or i minimum. Sometimes, we can determine their nature by simple observation of the equation , ;z,= f(x.y). In some cases we can apply the second derivative test, by eliminating the i ,dependentvariable. In fact, you can observe that equations (14), (15) and (16) are obtained by equating the partial i derivatives of the function ~~plicationsof Partial Derivatives to 0, treating x, y and h as independent variables. We can explain this in a simple way as follows : Suppose we are given the function f(x, y), whose extrema are to be found subject to the constraint g(x, y) = 0. We form the auxiliary function F(x, y, 1)= f(x, Y) + g (x, Y), ... (17) where h is to be determined. Then we find the three partial derivatives of F(x, y, h) and equate these to 0. Then we solve the three equations. The values of (x, y) thus obtained are the stationary points of the given function under the given constraint. The number h is called Lagrange's multiplier after Joseph Louis Lagrange, a leading mathematician of the 18th century. Here are some examples to illustrate the procedure. Lagrange (1736-1813) Example 18 :Let us find the largest and smallest values of f(x, y) = x + 2y on the circle x2+y2= 1. Fig. 6 In Fig. 6 you can see that f takes its maximum at a point in the first quadrant and its minimum .at a point in the third quadrant. Here f(x, y) = x + 2y and g(x, y) = x2+ y2 -1. .. The auxiliary functiod is Therefore, to find stationary points we have to solve the system of equations, Solving the first two equations, we get Taylor's Theorem Using the third equation, we get 1 The value I = 6gives 1 The value A = - 5 6yields. are 12 -1 -2 Thus, the stationary points ( -6'6 -) and (-6'6 -). 1 L I L Sincef (z. z) = 45 - -) = -4 5, we get that the largest value is 45 andf(-z, fi and the smallest value is -&. Example 19 : Suppose we want to find the extreme values of the function f (x, y) = xy on the surface g (x, y), where We first write the auxiliary function Now we have to solve the system of equations and x2 + 4y2= 8 i X2 From the first two equations we get x = - x , or X = f 2. k 4 Then x = f 2y. Substituting this in the third equation, we get 4~~+4y~=8=>~=f1. Correspondingly, we get x = f 2. Thus, the extreme values are obtained at the four points ! (2,1), (2, -1). (-2,l) and (-2, -1). The distinct values at these points are given by f(x,y) = 2 I and f(x, y) = -2. Therefore, the maximum value is 2 and the minimum value is -2. x2 NOR that f(x, y) = xy represents a hyperboloid and g(x. y) = - + YI - 1 = 0 represents an I 8 2 ellipse. In Figure 7, you can see the points of extrema off, subject to the condition that I g (x, Y).= 0. Example 20 : Let us find the right angled triangle of perimeter 1 with the largest area. Suppose ABC is a right angled triangle with perimeter 1 (See Fig. 8) Fig. 8 Let the sides of the triangle be x, y, dx2+ y2. Complete the rectangle ABCD with AC as 1 diagonal. Then 1 f(x, y) = area of A ABC = - x y. 2 The perimeter of the triangle = x + y + 4 x + y 2. We know that x+y+w=1. We have to find the maximum off (x, y) subject to the condition that g(x,y)=x+y+4x2+y2 =l. Let us form the system of equations for this f and g. We get Taylor's Theorem From the first two.equations we have or y-xy =x-xy Therefore, the sides are x, x, and & x, such thatx+x+ fix=1, i.e.,'(2+fi)x = 1. 1 1 Thus, the required sides are - and- 2+43'2+42 1+4Z The following exercises will give you enough practice of solving problems by using the method of Lagrange's multipliers. E20) Find the maximum value of f(x, y) = 3xy on 2x +y = 8. E21) Of all the pairs of numbers whose sum is 70, find the one that has the maximum product. E22) Find the minimum value of the function f(x, y) = x + y2 on 2x2 + y2 = 1. E23) Find the point on the parabola y - x2 = i) that maximizes the function f(x, y) = 2x -- y. Now let us quickly recall the points covered in this unit. 8.5 SUMMARY In this unit, we have 1) Defined Taylor polynomials of any order for functions of two variables. 2) Stated Taylor's theorem for functions of two variables. The second Taylor expansion: 3) ' Defined the terms 'local maxima' ,'local minima' and 'stationary point' for functions of two variables and discussed the relationship between them. Applications of Partial 4) Derived the second derivative test for points of extrema: Derivatives i) f (x, y) has a minimum (maximum) if b2 - 4 ac c 0 and a > 0 (a c 0). ii) f (x , y) has neither maximum nor minimum if b2 - 4 ac > 0. 5) Used the method of Lagrange multipliers to find the maximum and minimum values of a function of two variables. 8.6 SOLUTIONS AND ANSWERS El) f(x) = eK. The different derivatives off are f (x)=eX,f'(x)=e\ ..... ,fn (x)=ea Thus,f (2)=e2,f'(2)=e2,.... f" (2)=e2 Then the nth Taylor polynomial is given by e2(x-2) + e2 (x-2)' e2 (x-2)" P, (x) = e2+- + .... + l! 2! n! E2) f(x) = sin x. The different derivatives off are given by f(x) = sin x, f (x) = cos x fZ'(x) = -sin x, e3'(x) = - cos x f14' (x) = sin x, f5)(x) = cos x and f6)x = - sin x. Thus, f(0) = 0 ,f' (0) = 1, f;L (0) = 0, e3'(0) = -1, e4'(0) = 0, r5'(0) = 1 and r6'(x) = 0. This shows that the even derivatives are zero whereas the odd derivatives are alternatively + 1 and - 1. Thus, the 6th Taylor polynomial off is given by P, (x) = x - x3 + x5 E3) a) Let f(x) = x2- 3x + 4. Then and f' (x) = 2. f (-2) Therefore, f(-2) = 14, - -7, l! - Hence the polynomial is P2(x) = 14 - 7 (x + 2) + 1 (x + 2)* ) By Theorem 1, there exists a unique polynomial P(x) given by 'Taylor's Theorem 1 E5) Here f(x) = f(0) = 1 -1 +x' - 1 f (x) = (1 + x)~' The function f(x) and its derivatives of different orders are continuous in 1 ] - - , 1 [. Therefore, by Taylor's theorem 2 where c is a point between 0 and x. E6) We have calculated derivatives upto order 6 of the function f(x) = sin x in E2). That pattern shows that the derivatives of even order are alternatively sin x and - sin x whereas the derivatives of odd orders are alternatively - cos x and + cos x. Therefore, f(2k) (x) = (-l)k sin x, fiUL+"(x)= (- 1)' cos x Also the derivatives of any order are continuous and / f" (x) / 5 1 for any n and any X. Therefore, by Taylor's theorem, we have 5t where c is a point between - and x. 6 c) 4 E8) f(~,y)=e'+~ f(0,O) = 1 f, (x, Y) = fy(x, y) = fxy(x, y) = fxx(x, y) = fyy (x, y) =ex+Y. Thus, fx(0,O) = fy(0,O) = fx,(O, 0) = fxx(0,O) = fyy(0,O) = 1. Hence the 2nd Taylor polynomial P2(x, y) off is 33 I Appllcations df Partial Derivatives .. f., (x. Y) = 0 , fxy(1.1)=0 f,, (x, y) = 6x , f,,(l, 1) = 6 fyy (x, Y) = 6~ , fyy(1.1)=6. The Taylor polynomials are given by Po (x, y) = f (1, 1) = 4. P, (x, y) = 4 + 3(x -1) + 3 (y -1) = 3x + 3y - 2. 1 Pz (x, Y)= PI (x, y) + 5 [ 6 (X- I)'+ 6 (y -I)'] =3x+3y -2 +3(x - 112+3(y - 1)2 P, (x, y) = P2 (x, y) for m > 2. E10) Let P(x, y) = a, + a,,x + a,,, y + [ a,, x' + all xy + a,,, y21 P(O.0) = acg. dP -dn. =alo+2a20x +ally, Thus the 2nd Taylor polynomial of P is 1 T2(~,y)=%+a,ox+a,,ly -[2amx2+2al1dy+2a,,2y2] + 2 =~+alOx+rbly+ [amx2+allxy+rb2y21 = P(x. y) El 1) f(x, y) = xy2 + cos xy df - = 2xy - x sin xy, (df)- =n-1 ay JY (1,:) -a &2 = y2 cos xy, () = 0. '1. 71' a2f = 2y - [ sin xy + yx cos xy] dx dy Taylor's Theorem The second Taylor expansion is E12) f(x, y) = exsin y Then the required polynomial is = y + xy. E13) Suppose f is a function of two variables. We say that the function f has a local minimum at P(%, yo) if there exists some open disc S(P, r), r E R+,contained in the domain of f, such that for all (x, y) E S (P, r) we .have f(x, y) 2 f (xo , YO). df -- - 0 => 2xy = 0 => xy = 0. This shows that either x = 0 or y = 0. But dy y cannot be zero because y2 > 0 by Equation (*). Therefore x = 0. Then y2 = 4 and y = + 2 Hence the points are 10,2) and (0, -2). E15) f(x, y) = x2 + yZ- 8x - 2y + 18 is differentiable everywhere on the plane. Now check whether (4, 1) is a point of maxima or minima. Note that f(4, I)= 16+ 1 -32-2+ 18= 1 Now f(x, y) can be written as, f(x, y) = x2-8x + y2-2y + 18 =x2-8x+16+y2-2y+1 +1 'I = (~-4)~+(y - I)~+1 Applications of Partial 21=f(4, I) Derivatives Thus (4,l) is a global minimum for f. a) f(x, y) = 1 + x2 - y3 . Then The stationary points are the solutions of df df - = 2x = 0 and - - -3y2 = 0 dx *- . This happens if and only if x = 0 and y = 0. Thus (0,O) is the only stationary point off. b) f(x, y) = (X+ y) e-"'. Then =(-xy-x2+ l)e-"' Then the stationary points are solutions of -xy-y2+ 1 =Oand-xy-x2+ 1 =O Thus, This happens if and only if x2 = y? Therefore x = + y. But x = - y is not possible since otherwise 1 = xy + y2 = -y2 + y2 = 0. :. Substituting x= y in any of the earlier equations, we get that 1 y = f - . Thus the points are 4-2 a) The function f(x, y) = x2 + 2y2 - x has continuous partial derivatives of any df df order. everywhere on the plane. - = 2x - 1, dy = 4y. dx The stationary points are given by 2x - 1 = 0 and 4y = 0 1 i.e., x = - and y = 0. 2 1 Thus (- , 0) is the only stationary point off. Now, 2 This shows that bz - 4ac = -32 < 0 and a > 0. Therefore, by Theorem 7 f has Taylor's Theorem 1 I a minimum at (- , 0). 2 k 1 1 The extreme value is f (- ,0) = - - . 2 4 3y2 + 6xy = 0 ...... (**) 2 From (**),3y (y +2x)=O=> y =Oory =-2x ...... Substituting y = 0 in (*), we get that x = 1. Thus, (1, 0) is a stationary point. Now substituting y = - 2x in (*) , we get 1 -2 1 Thus, the stationary points are (1, , (- , - ) and (- , 1) b) 3 3 2. We first check the point (1.0) for extreme values. We get b2-4ac=-48<0anda> 0. Therefore the function has a local minimum at (1,O). The minimum value is f(1,O) = -1 12 Now at (- , - - ), 3 3 a = 2, b = -8, c = -2. b2-4ac=64+ 16>0anda> 0. 12 Therefore the function has no extremum at (- , - 3 ) . 3 1 f hasno extremum at (- - , 1) :. 2 C) f (x, y) = y + x sin y Applications of Partial sin y = 0 implies that y = 2nx or y = (2n + I) x , n is an integer. Then Derivatives cos((2n+l)x) =-1 andcos2nx= 1, sox = 1,orx =-1.Thus the stationary points are (1, 2nx ) and (-1 , (2n + 1) x ) &f 28f 8f = 0, -- '=2Cosy,- = -X sin y dX2 Ah dy ? 'Therefore, a = 0, b = + 2 , c = 0. and b2 - 4ac > 0 . Thus, f has no extremum at any point. f(x,y) = 2 cos (x + y) exY -df = -2 sin (X + y) + yeXY df df (0, 0) is a stationary point as both - and - vanish at (0,O). Now dx dy -2f = -2 cos (X + y) + y2 exy, (s)(o,O, = - 2 dX2 Then b2- 4ac = -12 < 0 and a < 0. Therefore, by Theorem 7, the function f has a local maximum at (0, 0). -df. =naxn-l -df = ncyn-1 dx 'dy df df - = 0 = - -> naxn-' = 0 and ncyn-' = 0. This is possible only when x = 0 dx dy - and y = 0, since a # 0 # c. Thus (0,O) is the only stationary point of f. Here we note that at (0, O), Therefore we cannot use Theorem 7. We have f (0,O) = 0. i) Now when a > 0 and c > 0, then f (x, y) = axn + cy" 0 = f (0, 0), if n is even. This shows that if n is even, then the function has a minimum at (Q, 0). The minimum value is 0. If n is odd, in every neighbourhood of (0.0) f takes on both positive and negative values f (x, y) < 0 for x < 0 and y < 0 , f (x, y) > 0 for x > 0 and y > 0. Therefore, if n is odd, then (0,O) is a saddle point off. . a% . ii) When a < 0, c < 0 , then f (x, y) = axn + cynI 0 = f(0, 0), if n is even. Therefore, if n is even,,f hasa maximum at (0.0). If n is odd, you can check that (0,O) is a saddle point. . Taylor's Theorem ' I - Now we have to solve the system of equations 3y+2A=O 2 1 Then y = - - h , x = - -A. Substituting in the third equation we get 3 1 3 Then x = 2, y = 4. f(2,4) = 24 Thus, (2,4) is an extremum. Now we shall check whether it is a maximum or minimum. The point (l,6) also satisfies g(x, y) = 0, and f(l.6)= 181f(2,4) Hence, f(2,4) = 24 is the maximum value off. Let x and y be the two numbers whose sum is 70 and whose product is maximum. Let f(x, y) = xy and g(x, y) = x + y. We have x + y = 70 . To maximize f(x, y) we use the method of Lagrange multipliers and solve the system of equations Then -21 = 70. Thus, A = -35. Therefore x = 35 and y = 35 and the extremum value is f (35,35) = 1225. Check that it is the maximum value. The system of equati'ons is I 2y = - h 2y => A = -1. Therefore x = - Substituting this in the third equation, 4 ' we get Applications of Partial The extreme value is given by Derivatives 17 . Check that fhis is the f(i, fi)= 2 + ,= 8 minimum value. E23) The point is (-1, 1).