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Math 1081 - Business Calculus Final Review Name: 1. A company wishes to manufacture a box with a volume of 36 ft3 that is open on top and is twice as long as it is wide. (a) Let x be the length of the box (so the width is 2x), and let y be the height of the box. First write the surface area of the box as a function of x and y. Then use the fact that the volume is 36 cubic feet to write the surface area as a function of x only. (b) Find the dimensions of the box that will lead to the minimum surface area. (c) What is the minimum surface area? 2. A fence must be built to enclose a rectangular area of 48 ft2 . Fencing material costs $4.00 per foot for the two sides facing north and south, and $3.00 per foot for the other two sides. (a) Let x be the length of the north and south sides of the fence, and let y be the length of the other two sides. First write the cost function for the fence in terms of x and y. Then use the fact that the area is 48 square feet to write the cost as a function of x only. (b) Find the dimensions of the fence that will lead to the minimum total cost. (c) What is the minimum total cost? 2 3. Find the following indefinite integrals. If you use integration by substitution, write what you chose to be u and find du. Don’t forget to add the constant! Z p (a) x 3x2 + 1 dx Z (4 + ln x)2 dx x Z x2 e6−x dx Z 2x3 − 3x dx x4 − 3x2 + 1 (b) (c) (d) 3 3 4. Evaluate the following definite integrals using the Fundamental Theorem of Calculus. Z 2 (3x3 − x + 5) dx (a) 0 Z 1 2 xex dx (b) −1 Z 0 (c) −3 1 dx (3 − x)2 4 5. Find fx (x, y) and fy (x, y) for the following functions. (a) f (x, y) = 4x2 y 3 + xy 5 − 6x (b) f (x, y) = yex (c) f (x, y) = 2y 3x + y 7xy − 2 (d) f (x, y) = ln |x8 + 3xy 2 | 6. Find the area bounded by the curves f (x) = x2 and f (x) = x7 on the interval [0, 1]. 5 Solutions 1. (a) SA = 2x2 + 6xy = 2x2 + 108 x (b) x = 3 feet, y = 2 feet (c) 54 square feet 2. (a) C = 8x + 6y = 8x + 288 x (b) $96 3. (a) (b) (c) (d) 3 1 2 2 9 (3x + 1) + c 1 3 3 (4 + ln x) + c 3 − 13 e6−x + c 1 4 2 2 ln |x − 3x + 1| +c 4. (a) 20 (b) 0 (c) 1 6 5. (a) fx (x, y) = 8xy 3 + y 5 − 6, fy (x, y) = 12x2 y 2 + 5xy 4 2 2 (b) fx (x, y) = 2xyex y , fy (x, y) = ex (x2 y + 1) 6. (c) fx (x, y) = −7y 2 − 6 −21x2 − 2 , f (x, y) = y (7xy − 2)2 (7xy − 2)2 (d) fx (x, y) = 8x7 + 3y 2 6xy , fy (x, y) = 8 8 2 x + 3xy x + 3xy 2 5 24 6