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M408C: Optimization Problems and Antiderivatives
October 14, 2008
Optimization problems are similar to related rates problems in that they both begin by needing to identify some equations. Once we have the equations, we want to find the absolute maxima and/or minima. If our function is defined on a closed interval, we can use the closed interval method. Otherwise, we use our work on determining how the derivative affects a function’s graph to find the max/min. Namely, we can use a variant of the First Derivative Test to find absolute max/min for functions defined on any interval. For antiderivatives, refer to p. 276, Table 2.
(4.7.39) A cone with height h is inscribed in a larger cone with height H so that its vertex is at the center of the base of the larger cone. Show that the inner cone has max volume when h = H/3.
Solution: It is helpful to draw a picture, so I suggest you do that first. We want to maximize the function V = 13 πr^2 h where h is the height of the cone and r is the radius of the base of the cone. This is a function of two variables, so we want to express one in terms of the other in order to get a function of one variable. From the picture, observe we have two similar right triangles. One has base r and height H − r (this one is above the top of the smaller cone) and the other has base R and height H (this one sits inside the big cone). Since these triangles are similar H−r h= HR. Thus r = (H − h)R/H. Plugging this in for V , we find
V (h) =
π(H − h)^2
R^2
H^2
· h = πR^2 3 H^2
(H^2 − 2 Hh + h^2 )h = πR^2 3 H^2
(H^2 h − 2 Hh^2 + h^3 ).
Note 0 ≤ h ≤ H, so we can apply the closed interval method to find the absolute max. To find the critical points, observe that
V ′(h) =
πR^2 3 H^2 (H^2 − 4 Hh + 3h^2 ) =
πR^2 3 H^2 (H − 3 h)(H − h).
Thus V ′(h) = 0 if and only if h = H/3 or h = H, one of the endpoints. Computing V at h = H/3,
we find that (after simplifying) V (H/3) = πR 2 3 H^2
4 H^3 27
- Checking the endpoints, we have that
V (0) = 0 and V (H) = 0. Thus the absolute max occurs at h = H/3 with volume πR^2 H/81.
- (4.9.30) Find f for f ′(x) = 2x − 3 /x^4 , x > 0, and f (1) = 3.
Solution: The antiderivative of f ′^ is f (x) = 2 · x 2 2 −^3
x−^3 − 3 +^ C^ =^ x
(^2) + x− (^3) + C. Since f (1) = 3,
we know that 1^2 + 1−^3 + C = 3, from which we conclude that C = 1. Hence f (x) = x^2 + x−^3 + 1.
- (4.5.32) Use the guidelines of Section 4.5 to graph y = x + cos x.
Solution: A, Domain: x ∈ (−∞, ∞). B, Intercepts: y(0) = 0 + cos 0 = 1 gives the y-intercept. For the x-intercept, note that y = 0 if and only if x + cos x = 0, i.e. cos x = −x. We don’t have the tools to solve this, so let’s skip it. C, Symmetry: y(−x) = −x + cos(−x) = −x + cos x which is neither y(x) nor −y(x). So there is no symmetry. D, Asymptotes: There are clearly no VAs. For HAs, note limx→∞ x+cos x = ∞ and limx→−∞ x+ cos x = −∞. So there are no HAs either. E, Intervals of Increase/Decrease: y′(x) = 1 − sin x, so y′(x) > 0 if and only if 1 > sin x, i.e. sin x < 1. But sin x < 1 for all x. Thus the function is always increasing. F, Local Max/Min: y′(x) = 0 if and only if sin x = 1. This happens at x = π 2 , − 32 π , 52 π , · · · = (4n+1)π
- These are the critical points, but since^ y
′ (^) is always positive, these are neither local max
nor local min. G, Concavity and IPs: y′′(x) = cos x, so y′′(x) > 0 if and only if cos x > 0. This happens for x ∈ (0, π/2) ∪ (3π/ 2 , 5 π/2) ∪ · · · , so y is concave up in these intervals. On the other hands, y′′(x) < 0 for x ∈ (π/ 2 , 3 π/2) ∪ (5π/ 2 , 7 π/2) ∪ · · · , so y is concave down at these intervals. H, Graph: You should be able to graph y, but see
http://www.ma.utexas.edu/~sbutt/teaching.html
for a link to a picture of the graph.
