Download Math 1081 - Business Calculus Final Review

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?
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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].
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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