Calculus: Optimization Worksheet

Question 1

Find two positive numbers whose sum is 40 and whose product is as large as possible.

Accepted Answer

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.

Question 2

A rectangle has perimeter 60 meters. Find the dimensions that maximize its area.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 4

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

Accepted Answer

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).

Question 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.

Accepted Answer

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.

Question 6

A closed cylindrical can must hold 500 cubic centimeters of liquid. Find the radius and height that minimize the surface area.

Accepted Answer

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.

Question 7

A right circular cone has volume 36pi cubic units. Find the radius and height that minimize its lateral surface area plus base area.

Accepted Answer

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)).

Question 8

Find the maximum value of f(x) = x^3 - 6x^2 + 9x + 2 on the interval [0, 5].

Accepted Answer

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.

Question 9

Find the minimum value of f(x) = x + 9/x for x > 0.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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?

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 13

Find the positive number x that minimizes the sum of x and the reciprocal of its square, f(x) = x + 1/x^2.

Accepted Answer

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).

Question 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?

Accepted Answer

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.

Question 15

Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius 5.

Accepted Answer

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.

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Calculus: Optimization

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