Calculus: Applications of Derivatives: Optimization Worksheet

Question 1

A rectangle has a perimeter of 40 meters. Find the dimensions that give the greatest possible area.

Accepted Answer

Let the sides be x and y. Since 2x + 2y = 40, y = 20 - x. The area is A = x(20 - x) = 20x - x^2, so A' = 20 - 2x. Setting A' = 0 gives x = 10, and then y = 10. The maximum area occurs for a 10 meter by 10 meter square.

Question 2

A 12 inch by 20 inch sheet of cardboard is used to make an open box by cutting equal squares of side length x from each corner and folding up the sides. Find the value of x that maximizes the volume.

Accepted Answer

The volume is V = x(12 - 2x)(20 - 2x), where 0 < x < 6. Expanding gives V = 240x - 64x^2 + 4x^3, so V' = 240 - 128x + 12x^2. Setting V' = 0 gives x = (16 - 2sqrt(19))/3, which is about 2.43 inches. This value gives the maximum volume because the other critical value is outside the domain.

Question 3

Find the point on the curve y = x^2 that is closest to the point (3, 0).

Accepted Answer

A point on the curve is (x, x^2). The squared distance to (3, 0) is D^2 = (x - 3)^2 + x^4. Its derivative is 2(x - 3) + 4x^3, so 4x^3 + 2x - 6 = 0. This has the solution x = 1, so the closest point is (1, 1).

Question 4

A farmer has 1200 feet of fencing to make a rectangular pen next to a straight river. No fence is needed along the river. Find the dimensions that maximize the enclosed area.

Accepted Answer

Let x be each side perpendicular to the river and y be the side parallel to the river. The constraint is 2x + y = 1200, so y = 1200 - 2x. The area is A = x(1200 - 2x) = 1200x - 2x^2. Since A' = 1200 - 4x, the critical point is x = 300, and y = 600. The maximum area is made by dimensions 300 feet by 600 feet.

Question 5

A company sells x items at a price of 100 - 2x dollars per item. The cost to produce x items is C(x) = 20x + 100. Find the number of items that maximizes profit.

Accepted Answer

Revenue is R(x) = x(100 - 2x) = 100x - 2x^2. Profit is P(x) = R(x) - C(x) = 80x - 2x^2 - 100. Since P'(x) = 80 - 4x, setting P'(x) = 0 gives x = 20. The profit is maximized when the company sells 20 items.

Question 6

A closed cylinder must have a volume of 500 cubic centimeters. Find the radius and height that minimize its surface area.

Accepted Answer

The volume constraint is pi r^2 h = 500, so h = 500/(pi r^2). The surface area is A = 2pi r^2 + 2pi rh = 2pi r^2 + 1000/r. Then A' = 4pi r - 1000/r^2. Setting A' = 0 gives r^3 = 250/pi, so r = (250/pi)^(1/3) and h = 2(250/pi)^(1/3). The minimum occurs when the height is twice the radius.

Question 7

A window is made from a rectangle topped by a semicircle. The total outside perimeter is 30 feet. Find the radius of the semicircle and the rectangle height that maximize the window area.

Accepted Answer

Let r be the semicircle radius and h be the rectangle height. The perimeter is 2h + 2r + pi r = 30, so h = (30 - (2 + pi)r)/2. The area is A = 2rh + (1/2)pi r^2, which simplifies to A = 30r - (2 + pi/2)r^2. Then A' = 30 - (4 + pi)r, so r = 30/(4 + pi). Substituting back gives h = 30/(4 + pi), so the rectangle height equals the semicircle radius.

Question 8

The height of a ball in feet after t seconds is s(t) = -16t^2 + 64t + 5. Find the maximum height and the time when it occurs.

Accepted Answer

The derivative is s'(t) = -32t + 64. Setting s'(t) = 0 gives t = 2 seconds. The height at that time is s(2) = -16(4) + 64(2) + 5 = 69. The maximum height is 69 feet after 2 seconds.

Question 9

Find the positive numbers x and y whose product is 36 and whose sum is as small as possible.

Accepted Answer

Since xy = 36, write y = 36/x for x > 0. The sum is S = x + 36/x. Then S' = 1 - 36/x^2. Setting S' = 0 gives x^2 = 36, so x = 6. Then y = 6, and the minimum sum is 12.

Question 10

Find the points on the parabola y = x^2 that are closest to the point (0, 4).

Accepted Answer

A point on the parabola is (x, x^2). The squared distance to (0, 4) is D^2 = x^2 + (x^2 - 4)^2. The derivative is 2x + 4x(x^2 - 4) = 2x(2x^2 - 7). The critical points are x = 0 and x = plus or minus sqrt(7/2). Checking distances shows the closest points are (sqrt(7/2), 7/2) and (-sqrt(7/2), 7/2).

Question 11

For f(x) = x^3 - 6x^2 + 9x + 4 on the interval 0 <= x <= 5, find the absolute maximum and absolute minimum values.

Accepted Answer

The derivative is f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3), so the critical points are x = 1 and x = 3. Check these and the endpoints: f(0) = 4, f(1) = 8, f(3) = 4, and f(5) = 24. The absolute maximum value is 24 at x = 5, and the absolute minimum value is 4 at x = 0 and x = 3.

Question 12

A vertical wall is 3 feet behind an 8 foot fence. A ladder must reach from the ground, over the top of the fence, to the wall. Find the shortest possible ladder length.

Accepted Answer

This classic optimization setup gives the minimum ladder length L = (3^(2/3) + 8^(2/3))^(3/2). Since 8^(2/3) = 4, the length is about (2.08 + 4)^(3/2), which is about 15.0 feet. The shortest possible ladder is approximately 15 feet long.

Question 13

A poster must contain 200 square inches of printed area. It has 2 inch side margins and 1 inch top and bottom margins. Find the printed dimensions that minimize the total poster area.

Accepted Answer

Let the printed area have width x and height y, so xy = 200 and y = 200/x. The poster dimensions are x + 4 by y + 2, so A = (x + 4)(200/x + 2) = 208 + 2x + 800/x. Then A' = 2 - 800/x^2. Setting A' = 0 gives x = 20, and y = 10. The printed dimensions should be 20 inches by 10 inches.

Question 14

A rectangular garden is built against a barn, so only three sides need fencing. The gardener has 90 meters of fencing. Find the maximum possible area.

Accepted Answer

Let x be each side perpendicular to the barn and y be the side parallel to the barn. The fencing constraint is 2x + y = 90, so y = 90 - 2x. The area is A = x(90 - 2x) = 90x - 2x^2. Since A' = 90 - 4x, x = 22.5 and y = 45. The maximum area is 22.5 times 45, which is 1012.5 square meters.

Question 15

An island is 6 kilometers offshore from the nearest point on a straight shoreline. A town is 10 kilometers down the shoreline from that nearest point. Cable costs 5 dollars per kilometer underwater and 3 dollars per kilometer on land. Find where the underwater cable should meet the shore to minimize total cost.

Accepted Answer

Let x be the distance along the shore from the nearest point to the island toward the town. The cost is C(x) = 5sqrt(x^2 + 36) + 3(10 - x), with 0 <= x <= 10. Then C'(x) = 5x/sqrt(x^2 + 36) - 3. Setting C'(x) = 0 gives 5x = 3sqrt(x^2 + 36), so x = 4.5. The cable should meet the shore 4.5 kilometers from the nearest shoreline point toward the town.

Calculus: Applications of Derivatives: Optimization

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