Practice Problems
Work through the odd-numbered problems 1-33. Once you have completed the problem set, check your answers.
Answers
1. (a) so . Maximize . and when . so yields the maximum enclosed area. When square feet.
(b) so . Maximize and when (then ). so yields the maximum enclosed area.
This garden is a by square.
(c) so . Maximize and when (then )
(d) A circle. A semicircle.
Fig. 3.5P1
3. (a) so . Maximize . and when (then ). so yields the maximum enclosed area. Area is square feet.
(b) A circular pen divided into equal stalls by two diameters shown in diagram (a) does a better job than a square with square feet. If the radius is , then so .
The resulting enclosed area is .
The pen shown in diagram (b) does even better. If each semicircle has radius , then the figure uses feet of fence so . The resulting enclosed area is
5. so
Maximize
and when and . When , then , clearly not a maximum, so . The dimensions of the box with the largest volume are , and .
7. (a) so .
and when (then )
(b) Let top bottom rate the bottom rate . Minimize . and when . If , then
If , then . If , then . As the cost of the bottom material increases, the radius of the least expensive cylindrical can decreases: the least expensive can becomes narrower and taller
9. Time distance/rate. Run distance ( Why?) so run time .
Swim distance so swim time and the total time is
when . The value so the least total time occurs when meters. In this situation, the lifeguard should run about of the way along the beach before going into the water.
11. (a) Consider a similar problem with a new town located at the "mirror image" of across the river (Fig. 3.5P11a). If the water works is built at any location along the river, then the distances are the same from to and to : dist .
Then . The shortest distance from to is a straight line (Fig. 3.5P11b), and this straight line gives similar triangles with equal side ratios: so miles. A consequence of this "mirror image" view of the problem is that "at the best location the angle of incidence equals the angle of reflection
(b) Minimize . so when and miles.
As it becomes relatively more expensive to build the pipe from a point on the river to , the cheapest route tends to shorten the distance from to .
Fig. 3.5P11
13. (a) Let be the length of one edge of the square end. Then so when or . The dimensions of the greatest volume acceptable box with a square end are by by inches: ,
(b) Let be the length of the shorter edge of the end. Then so when or . The dimensions of the largest box acceptable box with this shape are by by inches: .
(c) Let be the radius of the circular end. Then . so when or inches. The dimensions of the largest box acceptable box with a circular end are a radius of and a length of inches: .
15. Without calculus: The area of the triangle is (base)(height) (height) and the height is maximum when the angle between the sides is a right angle.
Using calculus: Let be the angle between the sides. Then the area of the triangle is
(base) (height) (height) . so
when , and the triangle is a right triangle with sides and .
Using either approach, the maximum area of the triangle is square inches, and the other side is the hypotenuse with length inches.
17. (a) . Then so when . The dimensions are and .
(b) . Then so when . The dimensions are and .
(c) The graph of is a "diamond" (a square) with corners at and . For so . Then and when . so we have a local . The dimensions are and .
(d) . Then so when . The dimensions are and .
19. and this is a maximum when . Then the maximum area is square inches. (This problem is similar to problem 15.)
21. and so . Then , and when so inches and inches.
23. Let be the number of passengers. The income is . The cost is so the profit is Income Cost .
and when passengers on the flight maximize your profit. (This is an example of treating a naturally discrete variable, the number of passengers, as a continuous variable.)
25. Apply the result of problem 24 with and .
27. (i) Let diameter of the base of the can, and let the height of the can.
Then .
For this can, and (sorry this should be in the statement of the problem) so which is about . The can can be tilted about before it falls over.
(ii) so when so : the height of the cola is .
(iii) (The center of gravity is exactly at the top edge of the cola. It turns out that when the of a can and liquid system is as low as possible then the cg is at the top edge of the liquid.) Then (radius/(height of ) which is about . In this situation, the can can be tilted about before it falls over.
(iv) Less.
29. (a) A (base)(height) = for . if .
(Clearly the endpoints and will not give the largest area.) Then .
(b) A (base)(height) = for
if . Then .
(c) A=(base)(height) = for .
if . Then .
31. (a) = (base)height) = . when . Then and Area = .
(b) . = (base)height) = . when . Then and Area = .
33. = = (top cost) + (bottom) + (sides) = But we know so so .
Then when . Now you can find and .