Sect. 15.7 (Min/Max/Saddle) —MATH 2421

Kawai

Sect. 15.7 (Min/Max/Saddle) 1. Single-variable functions:

(a) In principle, a constant function (horizontal line piece) has many local minimum and maximum points, but this is not interesting. We REALLY want: If f has a local maximum at x = a, then f (x) < f (a) for all x in some (small) interval around x = a. Often, we write the open interval this way: (a δ, a + δ) , where δ could be a really small positive number. − Thus, a local maximum should really be at the top of a hill, or, theoretically, it could be an isolated point: 1, x = 0 f (x) = 0, x = 0 6 We note that there is no mention of differentiability or even continuity in the definition of local maximum. The definition is similar for a local minimum.

(b) If f is diﬀerentiable (and therefore continuous), then all local extrema must have f 0 (a) = 0, a horizontal tangent line.

The converse is not true. If f 0 (a) = 0, then x = a could be an inﬂectionpoint. [Example: f (x) = x3 at x = 0.]

If x = a is a critical number of f, then either f 0 (a) = 0 or f 0 (a) is undeﬁned. This takes care of sharp turns (cusps) and jump discontinuities.

(c) If f 00 (a) exists and is not equal to zero, then the curve is concave up when f 00 (a) > 0 and concave down when f 00 (a) < 0. It is inconclusive when f 00 (a) = 0. If the concavity changes at x = a, then x = a is definitely an inflection point. However, we see that f (x) = x4 must have a local minimum at x = 0 since the function is nonnegative, but f 00 (0) = 0 because the bottom of the bowl is too flat.

(d) Second Derivative Test: If x = a is a critical number via f 0 (a) = 0, then the second derivative (if it exists) may classify the points as an extremum or an inﬂection point.

As in the previous bullet, if the concavity changes at x = a, then we must have f 00 (a) = 0, and this is an inflection point. If the curve is concave up in the neighborhood around x = a, then x = a must be a local minimum. Thus, if f 00 (a) > 0, then we know with certainty that x = a is a local minimum. However, this definition also tells us what to examine when f 00 (a) = 0 and the curve is very flat x = a. Similarly, if the curve is concave down in that neighborhood, then x = a must be a local maximum.

(e) If f 00 (a) = 0 then it an inﬂection point if and only if the concavity changes! As we have seen above, if the concavity does not change and f 0 (a) = 0, then it is a local extremum.

1 2. For a 2-var. function, z = f (x, y) , we say that a point (x0, y0) is a critical point (not a critical number) if: fx (x0, y0) = 0 AND fy (x0, y0) = 0 OR either partial derivative is undeﬁned. In the latter case, this could be due to a sharp turn in the surface or a vertical tangent line/plane, or a jump discontinuity.

3. The Second Partials Test is similar to the Calc. 1 Second Derivative Test. Recall that in Calc. 3, for a 2-var. function, there are FOUR 2nd partial derivatives: fxx, fxy, fyx, fyy.

From Clairaut, we know that fxy = fyx, so there are only three partial derivatives that we need to calculate for the test. From the Calc. 2 Maclaurin series, we recall that if we have an infinitely differentiable function at x = 0, then we can find the “best”quadratic (parabola) approximation to y = f (x) using this polynomial: f (0) f (0) p (x) = f (0) + 0 x + 00 x2. 1! 2!

This assumes that f 00 (0) = 0 and we want the best parabola around the point (0, f (0)) . When we move away from x =6 0, then the approximation ﬁts poorly, in general. Amazingly, this idea can be extended to 2nd degree polynomials in x and y. The Calc. 3 “best”quadric approximation to z = f (x, y) near (x, y) = (0, 0) is:

f (0, 0) f (0, 0) f (0, 0) f (0, 0) f (0, 0) z = f (x, y) f (0, 0)+ x x+ y y+ xx x2+ xy xy+ yy y2 ≈ 1! ∗ 1! ∗ 2! ∗ 1! 1! ∗ 2! ∗ ∗

If we assume that fx (0, 0) = fy (0, 0) = 0, then (0, 0) is a critical point and we have:

f (0, 0) f (0, 0) f (0, 0) z = f (0, 0) + xx x2 + xy xy + yy y2 2! ∗ 1! 1! ∗ 2! ∗ ∗ and this can only be one of two quadric surfaces. It is either an elliptic paraboloid, in which case, (0, 0) is either a local min. or max. Else, it must be a hyperbold paraboloid, in which case, (0, 0) is a saddle point. [A saddle point has at least one direction that goes up, and at least one direction that goes down. Thus, it cannot be a local extremum.]

4. Example 1: Suppose we have z = x2 + xy + y2. We are sure that we have a critical point at (0, 0):

fx = 2x + y = 0

fy = x + 2y = 0

and that we have a unique solution by substitution.

2 If we can get a graph of the level curve at k = 1, then we have:

x2 + xy + y2 = 1

2 y It’san ellipse! 1 Since this is a quadric surface, it must be an elliptic paraboloid and (0, 0) is either a local min. or local max. •2 •1 1 2 x •1

•2

What if we could not produce this contour? (Sadness.) We can look at the 2nd partial derivatives and ﬁnd:

fxx = 2, fxy = 1. and fyy = 2.

5. The proof of the 2nd Partials Test in Appendix B is probably not very convincing, so I present another way to look at this. If you don’t know how to multiply matrices, then don’t worry. This idea is presented to students who have some advanced background in linear algebra. If we have an ellipse or hyperbola of the form:

Ax2 + Bxy + Cy2 = 1,

then let’sassume that both A and C are positive. It turns out that if A and C have opposite signs, then it must be a hyperbola (diﬀerent algebra for that). If A and C are positive, then the relative size of B will determine which conic section this is. Matrix stuﬀ: This is called a quadratic form. AB/2 x x y = Ax2 + Bxy + Cy2 = 1 B/2 C y By multiplying the vectors on the sides of the 2 2 matrix, we get a 1 1 scalar polynomial. × × So the trick is to DIAGONALIZE/rotate the 2 2 matrix so that it ends up looking like this (a diagonal matrix): × P 0 0 Q For our matrix above, we want to diagonalize: 1 1/2 1/2 1 and the following rotation matrix does the trick:

√2/2 √2/2 − rotates points in the θ = +45◦ direction. √2/2 √2/2 3 Here is the diagonalization:

√2/2 √2/2 1 1/2 √2/2 √2/2 3/2 0 − = √2/2 √2/2 1/2 1 √2/2 √2/2 0 1/2 − In a new set of xy-coordinates, the equation of the ellipse becomes: 3 1 x2 + y2 = 1 and this must be an ellipse. 2 2

6. The Second Partials Test simpliﬁes this decision greatly. It turns out that we can determine the signs of the diagonal elements. Notice in the above example, the SUM of the diagonal elements in our original matrix was 2, and the sum of the diagonal elements in our rotated matrix is also 2. This is called the trace, and this value is invariant (does not change) under rotation. Thus, after our rotation, if both numbers are positive, then the critical point must be a local min. because it is a concave up elliptic paraboloid. If both numbers are negative, then the critical point must be a local max. If the numbers have opposite sign, then the critical point must be a saddle point, since the quadric surface is a hyperbolic paraboloid.

7. Here is the actual test:

(a) We assume that (x0, y0) is a critical point with fx = fy = 0.

(b) Evaluate fxx (x0, y0) , fyy (x0, y0) , and fxy (x0, y0) . (c) Let the discriminant:

2 d = fxxfyy (fxy) for that critical point. −

If d > 0, and fxx & fyy are positive, then it must be local min. If d > 0, and fxx & fyy are negative, then it must be local min. If d < 0, then it must be a saddle point. If d = 0, then the test is inconclusive (in a similar way that the Calc. 1 Second Derivative Test can be inconclusive).

8. Example 2: Suppose we have the function from p. 942:

f (x, y) = x (x 2) y (y + 3) − ∗ (a) Critical points. Again, we assume that we do not have any critical points where either fx or fy are undeﬁned. The ﬁrst condition gives us:

fx = 0 = 2y (y + 3) (x 1) x = 1 OR y = 0 OR y = 3. − ⇒ − Similarly, we have

fy = 0 = x (x 2) (2y + 3) x = 0 OR x = 2 OR y = 3/2. − ⇒ − 4 Now we must pair these conditions (there are 9 possible pairs taking one from fx = 0 and one from fy = 0). Only FIVE of the pairs make sense. For example, we can’t have x = 1 (from fx = 0) and x = 0 (from fy = 0) simultaneously. The ﬁve critical points are listed in the next part. (b) Now apply the Second Partials Test.

fxx = 2y (y + 3) , fxy = 2 (x 1) (2y + 3) − fyy = 2x (x 2) , fyx = fxy [Clairaut] −

2 Crit. pt. fxx fyy fxy d = fxxfyy (fxy) Result −

(0, 0) 0 0 6 36 Saddle pt. − −

(2, 0) 0 0 6 36 Saddle pt. −

(1, 3/2) 9/2 2 0 +9 Local Max. − − −

(0, 3) 0 0 6 36 Saddle pt. − −

(2, 3) 0 0 6 36 Saddle pt. − − −

9. Inconclusive??? When d = 0, the Second Partials Test is inconclusive. This could happen in diﬀerent ways, but here are the two semi-obvious ways... First, we know that in Calc. I, f (x) = x4 has a minimum at x = 0.

3 2 f 0 (x) = 4x . f 00 (x) = 12x .

Clearly, we have a critical number x = 0, but we cannot use the Second Derivative Test because f 00 (0) = 0.

The concavity is positive on both sides of x = 0 and we see that f 0 changes from negative to positive when passing through x = 0, so it must be a local min. The curve is too ﬂataround x = 0, and thus, the Second Derivative Test is inconclusive there.

5 The same thing happens when we look at Calc. III’s:

4 4 3 3 f (x, y) = x + y fx = 4x and fy = 4y . ⇒

The only critical point is (0, 0) , but fxx, fyy, and fxy are all zero there. From algebra, we know that (0, 0) must be a local min. point. Same issue. The surface is too ﬂat around the point (0, 0) , and this causes the Second Partials Test to be inconclusive.

Another interesting case:

f (x, y) = (x y)2 − fx = 2 (x y) and fy = 2 (x y) . − − − fxx = 2, fyy = 2, fxy = 2 d = 0. − ⇒ This gives us an inﬁnite number of critical points along the line y = x.

This is a parabolic cylinder that has been rotated 2 45◦ from z = y . Clearly, on the line y = x, we have a line of minimum points.

Second Partials Test does not like linked critical points!!

10. Addendum. Quadratic Formula for x2 xy + y2 1 = 0 Ellipse. − − ⇒ Solve for y: y2 xy + x2 1 = 0 − −

2 x ( x) 4 (x2 + 1) x √4 3x2 y = ± − − = ± − q 2 2

We see that the square root is only real when 4 3x2 0 x2 4/3. − ≥ ⇒ ≤ This is the interval 2/√3, 2√3 and, thus, it must be an ellipse. − A hyperbola would have an inﬁnite domain.