Week #7 – Maxima and Minima, Concavity, Applications Section 4.3

SUGGESTED PROBLEMS

1. Note that you can deﬁne the end points as local minima as well, if you like. We will accept both answers (end points are or are not local mins/maxes).

2. Note that you can deﬁne the end points as local minima as well, if you like. We will accept both answers (end points are or are not local mins/maxes). Note: the word “exact” is math-class code for “don’t give a rounded decimal expan- sion”. Your answer should be expressible using fractions, roots, etc. without resorting to decimals.

9. Diﬀerentiating using the product rule gives

′ −t −t −t g (t) = 1 · e − te = (1 − t)e

so the critical point is at t = 1. Since g′(t) > 0 for 0 1, the critical point is a local maximum. Now to determine if t = 1 is a global maximum: As t →∞, the value of g(t) → 0. Also, g(0) = 0, and g(t) is increasing until t = 1. Thus the local maximum at t =1 must be a global maximum. The global maximum value of g(t) for t> 0 is g(1) = 1 · e−1 = 1/e. A sketch of the graph is shown below.

1 g(t)

00 1 1/e t

11. Diﬀerentiating using the quotient rule gives

2 2 ′ 1(1 + t ) − t(2t) 1 − t f (t)= = (1 + t2)2 (1 + t2)2

The critical points are the solutions to

1 − t2 = 0 (1 + t2)2 t2 = 1 t = ±1

(There are no points where f ′(t) is undeﬁned, as the denominator is always positive.) Since f ′(t) > 0 for −1

−1/2 0−1 1/2 0 1 t

Thus t = 1 is a global maximum, and t = −1 is a global minimum.

17. Let y = ln(1 + x). Since y′ = 1/(1 + x), y is increasing for all x ≥ 0. The lower bound is at x = 0, or at a value of y(0) = ln(1) = 0. Thus ln(1 + x) ≥ 0 for x ≥ 0. From our knowledge of the graph of ln(x), we know there is no upper bound on ln(1+ x).

2 35. (a) From left to right, the graph of g(x) starts “flat”, decreases slowly at first then more rapidly, most rapidly at x = 0. The graph then continues to decrease but less and less rapidly until flat again at x = 2. The function has an even derivative, so the original function will be symmetric about the point (0, g(0)). A possible sketch of the graph is shown below.

∆1

∆2

−2 0 2

(b) The graph has an inﬂection point at (0, g(0)) where the slope changes from negative and decreasing to negative and increasing. (c) The function has a global maximum at x = −2 and a global minimum at x = 2. (d) Since the function is decreasing over the interval −2 ≤ x ≤ 2

g(−2) > g(0) > g(2) Since the function appears symmetric about (0, g(0)), the distance in the y direction from g(−2) to g(0) is equal to the distance from g(−2) to g(0). In the diagram,

∆1 =∆2

QUIZ PREPARATION PROBLEMS

19. We want to ﬁnd the maximum value of y. To ﬁnd the critical points, we need y′:

′ y = −32t + 50 Setting y’ = 0, 0 = −32t + 50 50 t = ≈ 1.56 seconds 32

Thus, we have the time at which the height is a maximum. The actual height at that point is

50 y( ) ≈ 44.1 feet 32

21. (a) We have

C D CD2 D3 T (D)= − D2 = − 2 3 2 3

3 and so dT = CD − D2 = D(C − D) dD dT/dD is zero when D =0 or D = C. We can also compute the sign of dT/dD:

Sign of T’(x)

0 0 − + −

0 C D So by the ﬁrst derivative test, D = 0 is a local min, and D = C is a local max. (b) The sensitivity is dT/dD, so it will be maximized when its derivative, d2T/dD2 equals zero.

d2T = C − 2D dD2 d2T Setting = 0, 0= C − 2D dD2 D = C/2

By the ﬁrst derivative test on d2T/dD2, the is critical point is a local maximum of dT/dD.

23. (a) Since y must be positive, the rate is non-negative (≥ 0) for (a − y) ≥ 0. The graph is a quadratic function with roots at y = 0 and y = a. Max Rate Rate 0 0 a/2 a y

(b) The Maximum value of the rate occurs at y = a/2. You can get this from the symmetry of a quadratic function (max is half-way between the two roots at y = a and y = 0, or by setting the derivative equal to zero. Because the rate is a quadratic function, the one local maximum will also be a global maximum.

25. (a) To show that R is an increasing function of r1, we show that dR/dr1 > 0 for all values of r1. We solve ﬁrst for R:

4 1 1 1 = + R r1 r2 1 r + r = 2 1 R r1r2 r r R = 1 2 r1 + r2

Now diﬀerentiate using the quotient rule. Treat r1 as the variable, and r2 as a constant.

2 dR (r1 + r2)(r2) − (r1r2)(1) (r2) = 2 = 2 dr1 (r1 + r2) (r1 + r2)

Since dR/dr1 is the quotient of two squared terms, it must always be positive. This means that R increases as r1 increases for all r1.

(b) Since R is increasing on any interval a ≤ r1 ≤ b, the maximum value of R occurs at the right endpoint r1 = b. 26. (a) We want the maximum value of I. Using the properties of logarithms, we rewrite the expression for I as

I = k(ln S − ln S0) − S + S0 + I0

Since k and S0 are constant, diﬀerentiating with respect to S gives dI k = − 1 dS S Thus the critical point is at S = k. Since dI/dS is positive for S < k and dI/dS is negative for S > k, we see that S = k is a local maximum. We only consider positive values of S. Since S = k is the only critical point, it gives the global maximum value for I, which is

I = k(ln k − ln S0) − k + S0 + I0

(b) Since both k and S0 are in the expression for the maximum value of I, both the particular disease and how it starts inﬂuence the maximum. 27. (a) For a point (t, s), the line from the origin has rise = s and run = t. Thus the slope of the line OP is s/t. This is the same as the chipmunk problem from the lecture notes.

5 (b) Sketching several lines from the origin to points on the curve, we see that the maximum slope occurs at the point Q, where the line to the origin is tangent to the graph. Reading from the graph, we see t ≈ 2 hours at this point.

(c) The instantaneous speed of the cyclist at any time is given by the slope of the corresponding point on the curve. At the optimal point Q, the line from the origin is tangent to the curve, so the quantity s/t equals the cyclist’s speed at the point Q.

33. (a) To obtain g(v), which is in gallons per mile, we need to divide f(b) (in gallons per hour) by v (in miles per hour). Thus, g(v)= f(v)/v. (b) By inspecting the graph we see that f(v) is minimized at approximately 220 mph. (c) Note that a point on the graph of f(v) has the coordinates (v,f(v)). The line passing through this point and the origin (0, 0) has

f(v) − 0 f(v) Slope = = = g(v) v − 0 v

So minimizing g(v) corresponds to ﬁnding the line of minimum slope from the family of lines which pass through the origin (0, 0) and the point (v,f(v)) on the graph of f(v). This line is the unique member of the family which is tangent to the graph of f(v). The value of v corresponding to the point of tangency will minimize g(v). This value of v will satisfy f(v)/v = f ′(v). From the graph below, we see that v ≈ 300mph

(d) The pilot’s goal with regard to f(v) and g(v) would depend on the purpose of the flight, and might even vary within a given flight. For example, if the mission involved aerial surveillance or banner-towing over some limited area, or if the plane was flying a holding pattern, then the pilot would want to minimize f(v) so as to remain aloft as long as possible. In a more common situation, where the purpose was economical

6 travel between two ﬁxed points, then the minimum net fuel expenditure for the trip would result from minimizing g(v) (i.e. since the trip is ﬁxed, minimize the fuel per mile).