To review these concepts, go to Calculus and Optimization Study Guide.
Calculus and Optimization Practice Problems
 Suppose a company makes a profit of makes and sells x > 0 items. How many items should it make to maximize profit?
 Suppose the profit of a company is when it makes items a day. What level of production will maximize profits?
 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?
 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?
 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?

A farmer has 400 feet of fencing to make three rectangular pens. What dimensions x and y will maximize the total area?

Four pens will be built along a river by using 150 feet of fencing. What dimensions will maximize the area of the pens?
 A rectangular pen will be built using 100 feet of fencing. What dimensions will maximize the area?
 The surface area of a can is Area = 2π r^{2} + 2π rh, where the height is h and the radius is r. The volume is volume = π r^{2}h. What dimensions minimize the surface area of a can with volume 16π cubic inches?
 A painter has enough paint to cover 600 square feet of area. What is the largest squarebottom box that could be painted (including the top, bottom, and all sides)?
 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?

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
 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.
 P'(x) = 18x + 40 – x^{2} = –(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.
 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 – 10x^{2}. 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.
 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.
 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.
 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.
 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 – 5y^{2}. 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.
 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) = 50x – x^{2}. 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.
 Because Volume = π r^{2}h = 16π, it follows that h = . Thus, the surface area function is Area(r) = 2 πr^{2}+ 2πr () = 2π + The derivative Area'(r) = 4πr – is zero when 4πr = , so r^{3} = 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.
 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 = x^{2}y and the surface area is Area = x^{2} + 4xy + x^{2} (the top, the four sides, and the bottom). And because Area = 2x^{2} + 4xy = 600, the height . Thus, the Volume = x^{2}y = x^{2}( – ) = 150x – x^{3}. The derivative Volume'(x) = 150 – x^{2} is zero when x^{2} = 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.
 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 = x^{2}, so it will cost x^{2} dollars to build. Similarly, it will cost 7x^{2} dollars to build the base. The total cost of the box is therefore Cost = 0.4xy + 8x^{2}. Because the volume is x^{2}y = 40,000, we know y = . Thus, the cost function can be written: Cost(x) = 0.4x() + 8x^{2} = + 8x^{2}. The derivative: Cost'(x) = – + 16x is zero only when x^{3} = 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.
 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 x^{2} = 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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