Calculus: Applications of Derivatives: Optimization
Using derivatives to maximize and minimize quantities
Calculus: Applications of Derivatives: Optimization
Using derivatives to maximize and minimize quantities
Math - Grade 9-12
- 1
A rectangle has a perimeter of 40 meters. Find the dimensions that give the greatest possible area.
Use the perimeter equation to write the area in terms of one variable.
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. - 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.
The height of the box is x, and the base dimensions are reduced by 2x in each direction.
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. - 3
Find the point on the curve y = x^2 that is closest to the point (3, 0).
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). - 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.
Only three sides need fencing.
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. - 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.
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. - 6
A closed cylinder must have a volume of 500 cubic centimeters. Find the radius and height that minimize its surface area.
Use the volume equation to replace h before differentiating surface area.
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. - 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.
The width of the rectangle is the diameter of the semicircle.
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. - 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.
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. - 9
Find the positive numbers x and y whose product is 36 and whose sum is as small as possible.
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. - 10
Find the points on the parabola y = x^2 that are closest to the point (0, 4).
Minimize the square of the distance instead of the distance itself.
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). - 11
For f(x) = x^3 - 6x^2 + 9x + 4 on the interval 0 <= x <= 5, find the absolute maximum and absolute minimum values.
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. - 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.
Use similar triangles to write the ladder length in terms of one ground distance, then differentiate.
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. - 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.
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. - 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.
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. - 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.
Use the Pythagorean theorem for the underwater cable length.
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.