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# Calculus and Optimization Practice Problems

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## Calculus and Optimization Practice Problems

1. Suppose a company makes a profit of makes and sells x > 0 items. How many items should it make to maximize profit?
2. Suppose the profit of a company is when it makes items a day. What level of production will maximize profits?
3. When 30 orange trees are planted on an acre, each will produce 500 oranges a year. For every additional orange tree planted, each tree will produce 10 fewer oranges. How many trees should be planted to maximize the yield?
4. An artist can sell 20 copies of a painting at \$100 each, but for each additional copy she makes, the value of each painting will go down by a dollar. Thus, if 22 copies are made, each will sell for \$98. How many copies should she make to maximize her sales?
5. A garden has 200 pounds of watermelons growing in it. Every day, the total amount of watermelon increases by 5 pounds. At the same time, the price per pound of watermelon goes down by 1¢. If the current price is 90¢ per pound, how much longer should the watermelons grow in order to fetch the highest price possible?
6. A farmer has 400 feet of fencing to make three rectangular pens. What dimensions x and y will maximize the total area?

7. Four pens will be built along a river by using 150 feet of fencing. What dimensions will maximize the area of the pens?

8. A rectangular pen will be built using 100 feet of fencing. What dimensions will maximize the area?
9. The surface area of a can is Area = 2π r2 + 2π rh, where the height is h and the radius is r. The volume is volume = π r2h. What dimensions minimize the surface area of a can with volume 16π cubic inches?
10. A painter has enough paint to cover 600 square feet of area. What is the largest square-bottom box that could be painted (including the top, bottom, and all sides)?
11. A box with a square bottom will be built to contain 40,000 cubic feet of grain. The sides of the box cost 10¢ per square foot to build, the roof costs \$1 per square foot to build, and the bottom will cost \$7 per square foot to build. What dimensions will minimize the building costs?
12. A printed page will have a total area of 96 square inches. The top and bottom margins will be 1 inch each, and the left and right margins will be inches each. What overall dimensions for the page will maximize the area of the space inside the margins?

### Solutions

1. P'(x) = + = 0 when x = 10. Using the first derivative test, P' (1) = 9,000 is positive, so the function increases to x = 10 and P'(100) = – + = – So the function decreases afterward, thus x = 10 maximizes the profit.
2. P'(x) = 18x + 40 – x2 = –(x – 20)(x + 2) is zero at x = 20 and at x = –2, although making a negative number of items is impossible. Thus, x = 20 is the only point of slope zero. The second derivative is P"(x) = 18 – 2x, which is negative when x = 20, thus this point is an absolute maximum according to the second derivative test.
3. If x is the number of trees beyond 30 that are planted on the acre, then the number of oranges produced will be: Oranges(x) = (number of trees) (yield per tree) = (30 + x)(500 – 10x) = 15,000 + 200x – 10x2. The derivative Oranges'(x) = 200 – 20x is zero when x = 10. Using the second derivative test, Oranges"(x) = – 20 is negative, so this is maximal. Thus, x = 10 more than 30 trees should be planted, for a total of 40 trees per acre.
4. The total sales will be figured as follows: Sales = (number of copies)(price per copy), so Sales(x) = (20 + x)(100 – x) where x is the number of copies beyond 20. The derivative Sales'(x) = 80 – 2x is zero when x = 40. Because Sales"(x) = –2, this is maximal by the second derivative test. Thus, the artist should make x = 40 more than 20 paintings, for a total of 60 paintings in order to maximize sales.
5. After x days, there will be 200 + 5x pounds of watermelon, which will valued at 90 – x cents per pound. Thus, the price after x days will be Price(x) = (200 + 5x)(90 – x) cents. The derivative is Price'(x) = 250 – 10x, which is zero when x = 25. Because Price'(x) is clearly positive when x is less than 25 and negative afterward, this is maximal by the first derivative test. Thus, the watermelons will fetch the highest price in 25 days.
6. The area is Area = xy and the total fencing is 4y + 2x = 400. Thus, x = 200 – 2y, so the area function can be written as follows: Area = xy = (200 – 2y) · y. The derivative Area'(y) = 200 – 4y is zero when y = 50. Because the second derivative is Area"(y) = –4, this is an absolute maximum. Thus, the optimal dimensions for the pen are y = 50 and x = 200 – 2y = 200 – 2(50) = 100.
7. Here, Area = xy and the total fencing is 5y + x = 150. Because x = 150 – 5y, the area function can be written as follows: Area(y) = (150 – 5y)y = 150y – 5y2. The derivative Area'(y) = 150 – 10y is zero when y = 15. Either the first or the second derivative test can be used to prove this is maximal. Thus, the optimal dimensions are y = 15 and therefore x = 150 – 5(15) = 75.
8. Suppose the height of the rectangle is y and the width is x. The area is thus Area = xy, and the perimeter is 2x + 2y = 100. Thus, y = 50 – x, so Area(x) = x(50 – x) = 50xx2. The derivative Area'(x) = 50 – 2x is zero when x = 25. The second derivative is Area"(x) = –2, so this is maximal by the second derivative test. The height y = 50 – x = 50 – 25 = 25 is the same as the width. Thus, the rectangle with the largest area for a given amount of perimeter is a square.
9. Because Volume = π r2h = 16π, it follows that h = . Thus, the surface area function is Area(r) = 2 πr2+ 2πr () = 2π + The derivative Area'(r) = 4πr is zero when 4πr = , so r3 = 8. Thus, the only point of slope zero is when r = 2. The second derivative is Area"(r) = 4π + , which is positive when r = 2. Thus, by the second derivative test, r = 2 is the absolute minimum. Thus, a radius of r = 2 and a height of h = = 4 will minimize the surface area.
10. Because the box has a square bottom, its length and width can be both x, while its height is y Thus, the volume is Volume = x2y and the surface area is Area = x2 + 4xy + x2 (the top, the four sides, and the bottom). And because Area = 2x2 + 4xy = 600, the height . Thus, the Volume = x2y = x2() = 150xx3. The derivative Volume'(x) = 150 – x2 is zero when x2 = 100. Negative lengths are impossible, therefore this is zero only when x = 10. By the second derivative test, Volume"(x) = –3x is negative when x = 10, so this is a maximum. The corresponding height is y = = 10, so the largest box is a cube with all sides of length 10.
11. Because the box has a square bottom, let x be both the length and width, and y be the height. The area of each side is thus xy, so the cost to build four of them at ten cents a square foot is 0.10( 4xy) = 04xy dollars. The area of the top is x · x = x2, so it will cost x2 dollars to build. Similarly, it will cost 7x2 dollars to build the base. The total cost of the box is therefore Cost = 0.4xy + 8x2. Because the volume is x2y = 40,000, we know y = . Thus, the cost function can be written: Cost(x) = 0.4x() + 8x2 = + 8x2. The derivative: Cost'(x) = – + 16x is zero only when x3 = 1,000 or x = 10. By the second derivative test, Cost"(x) = + 16 is positive when x = 10, so this is the absolute minimum. The cheapest box will be built when x = 10 and y = 400.
12. Inside the margins, the area is: Area = (x –2 (1))(y –2) = (x – 3)(y – 2) = xy – 2x – 3y + 6. The total area of the page is xy = 96, so y = . Therefore, Area(x) = x () – 2x – 3y + 6 = 102 – 2x. The derivative Area '(x) = –2 + is zero when x2 = 144, thus when x = 12 (ignore negative lengths). The second derivative Area"(x) = – is negative when x = 12, so this is the absolute maximum. Thus, the dimensions that maximize the printed area are x = 12 and y = 8.

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