Calculus: Optimization
Using derivatives to maximize and minimize quantities
Calculus: Optimization
Using derivatives to maximize and minimize quantities
Calculus - Grade advanced
- 1
Find two positive numbers whose sum is 40 and whose product is as large as possible.
Write the product using only one variable before differentiating.
Let the numbers be x and 40 - x. The product is P(x) = x(40 - x) = 40x - x^2. Since P'(x) = 40 - 2x, setting P'(x) = 0 gives x = 20. The numbers are 20 and 20, and their maximum product is 400. - 2
A rectangle has perimeter 60 meters. Find the dimensions that maximize its area.
Use the perimeter equation to solve for one side in terms of the other.
Let the length be x and the width be y. The constraint is 2x + 2y = 60, so y = 30 - x. The area is A(x) = x(30 - x) = 30x - x^2. Since A'(x) = 30 - 2x, the critical point is x = 15, so y = 15. The maximum area occurs when the rectangle is a 15 meter by 15 meter square. - 3
A farmer has 200 meters of fencing to make a rectangular pen along a straight river. The side along the river needs no fence. Find the dimensions that maximize the enclosed area.
Only three sides require fencing.
Let x be each side perpendicular to the river and y be the side parallel to the river that is fenced. The constraint is 2x + y = 200, so y = 200 - 2x. The area is A(x) = xy = x(200 - 2x) = 200x - 2x^2. Since A'(x) = 200 - 4x, the critical point is x = 50. Then y = 100. The maximum area is 5000 square meters. - 4
Find the point on the parabola y = x^2 that is closest to the point (0, 4).
Minimize the square of the distance instead of the distance itself.
A point on the parabola has coordinates (x, x^2). The squared distance to (0, 4) is D(x) = x^2 + (x^2 - 4)^2. Differentiating gives D'(x) = 2x + 4x(x^2 - 4) = 2x(2x^2 - 7). Thus x = 0 or x = ±sqrt(7/2). The distances squared are 16 at x = 0 and 7/2 + (-1/2)^2 = 15/4 at x = ±sqrt(7/2). The closest points are (sqrt(7/2), 7/2) and (-sqrt(7/2), 7/2). - 5
An open-top box is made by cutting equal squares of side length x from the corners of a 30 cm by 20 cm sheet of cardboard and folding up the sides. Find the value of x that maximizes the volume.
The height is x, and each original side loses 2x after the corner squares are removed.
The box dimensions are length 30 - 2x, width 20 - 2x, and height x. The volume is V(x) = x(30 - 2x)(20 - 2x), where 0 < x < 10. Expanding gives V(x) = 600x - 100x^2 + 4x^3. Then V'(x) = 600 - 200x + 12x^2. Setting V'(x) = 0 gives 3x^2 - 50x + 150 = 0, so x = (25 ± 5sqrt(7))/3. Only x = (25 - 5sqrt(7))/3 is in the domain, which is about 3.92 cm. This value maximizes the volume. - 6
A closed cylindrical can must hold 500 cubic centimeters of liquid. Find the radius and height that minimize the surface area.
For a closed cylinder, include both circular ends in the surface area.
The volume constraint is pi r^2 h = 500, so h = 500/(pi r^2). The surface area is S = 2pi r^2 + 2pi r h. Substituting gives S(r) = 2pi r^2 + 1000/r. Then S'(r) = 4pi r - 1000/r^2. Setting S'(r) = 0 gives 4pi r^3 = 1000, so r = cuberoot(250/pi). The height is h = 500/(pi r^2), which simplifies to h = 2r. Thus the minimizing dimensions satisfy r = cuberoot(250/pi) cm and h = 2cuberoot(250/pi) cm. - 7
A right circular cone has volume 36pi cubic units. Find the radius and height that minimize its lateral surface area plus base area.
The slant height is sqrt(r^2 + h^2), and the cone has one circular base.
The volume constraint is (1/3)pi r^2 h = 36pi, so h = 108/r^2. The surface area is S = pi r^2 + pi r sqrt(r^2 + h^2). Substituting h = 108/r^2 gives a one-variable function. A cleaner method uses the known condition for fixed volume that total surface area of a cone is minimized when h = sqrt(2)r. Then (1/3)pi r^2(sqrt(2)r) = 36pi, so r^3 = 108/sqrt(2) = 54sqrt(2). Therefore r = cuberoot(54sqrt(2)) and h = sqrt(2)cuberoot(54sqrt(2)). - 8
Find the maximum value of f(x) = x^3 - 6x^2 + 9x + 2 on the interval [0, 5].
For a closed interval, compare values at critical points and endpoints.
Differentiate to get f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3). The critical points in the interval are x = 1 and x = 3. Evaluate the function at endpoints and critical points: f(0) = 2, f(1) = 6, f(3) = 2, and f(5) = 12. The maximum value on [0, 5] is 12, which occurs at x = 5. - 9
Find the minimum value of f(x) = x + 9/x for x > 0.
The derivative is f'(x) = 1 - 9/x^2. Setting f'(x) = 0 gives x^2 = 9, so x = 3 because x > 0. Since f''(x) = 18/x^3 is positive for x > 0, this point gives a minimum. The minimum value is f(3) = 3 + 3 = 6. - 10
A company finds that the profit from selling x hundred items is P(x) = -2x^3 + 24x^2 + 90x - 50, where 0 <= x <= 10. Find the production level that maximizes profit.
Remember that x is measured in hundreds of items.
Differentiate to get P'(x) = -6x^2 + 48x + 90. Setting P'(x) = 0 gives -6(x^2 - 8x - 15) = 0, so x = 4 ± sqrt(31). Only x = 4 + sqrt(31) is in [0, 10]. Comparing with endpoints confirms this critical point gives the maximum. The company should produce 4 + sqrt(31) hundred items, or about 957 items. - 11
A wire 100 cm long is cut into two pieces. One piece is bent into a square, and the other is bent into a circle. How should the wire be cut to minimize the total enclosed area?
Write each area in terms of the length of wire used to form that shape.
Let x be the length used for the square, so 100 - x is used for the circle. The square has side x/4 and area x^2/16. The circle has circumference 100 - x, radius (100 - x)/(2pi), and area (100 - x)^2/(4pi). The total area is A(x) = x^2/16 + (100 - x)^2/(4pi). Then A'(x) = x/8 - (100 - x)/(2pi). Setting A'(x) = 0 gives pi x = 4(100 - x), so x = 400/(pi + 4). Thus 400/(pi + 4) cm should be used for the square and 100pi/(pi + 4) cm should be used for the circle. - 12
A rectangle is inscribed under the curve y = 12 - x^2 and above the x-axis, with its base on the x-axis and its sides vertical. Find the dimensions of the rectangle with maximum area.
Use symmetry to describe the rectangle with one positive variable x.
By symmetry, let the upper corners be at (-x, 12 - x^2) and (x, 12 - x^2), where 0 < x < sqrt(12). The width is 2x and the height is 12 - x^2. The area is A(x) = 2x(12 - x^2) = 24x - 2x^3. Then A'(x) = 24 - 6x^2. Setting A'(x) = 0 gives x = 2. The maximum rectangle has width 4 and height 8. - 13
Find the positive number x that minimizes the sum of x and the reciprocal of its square, f(x) = x + 1/x^2.
For x > 0, f'(x) = 1 - 2/x^3. Setting f'(x) = 0 gives x^3 = 2, so x = cuberoot(2). Since f''(x) = 6/x^4 is positive for x > 0, this critical point gives a minimum. The minimizing value is x = cuberoot(2). - 14
A boat is 3 km from the nearest point on a straight shoreline. A destination is 10 km along the shoreline from that nearest point. The boat travels at 6 km/h on water and 10 km/h on land. At what point along the shore should the boat land to minimize total travel time?
Minimize time, not distance. Use time equals distance divided by speed.
Let x be the distance along the shore from the nearest point to the landing point, with 0 <= x <= 10. The water distance is sqrt(x^2 + 9), and the land distance is 10 - x. The time is T(x) = sqrt(x^2 + 9)/6 + (10 - x)/10. Then T'(x) = x/(6sqrt(x^2 + 9)) - 1/10. Setting T'(x) = 0 gives 10x = 6sqrt(x^2 + 9). Squaring gives 100x^2 = 36x^2 + 324, so 64x^2 = 324 and x = 9/4. The boat should land 2.25 km along the shore from the nearest point. - 15
Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius 5.
Use the circle equation for a corner of the rectangle.
Let the rectangle have half-width x and half-height y. Its corners lie on the circle x^2 + y^2 = 25, and its area is A = 4xy. Solving for y gives y = sqrt(25 - x^2), so A(x) = 4x sqrt(25 - x^2). The maximum occurs when x = y by symmetry or by differentiating. Thus x = y = 5/sqrt(2), so the rectangle has width 10/sqrt(2) = 5sqrt(2) and height 5sqrt(2). The largest rectangle is a square.