MGSC: Cube Centimeters Test

Topics: Marginal cost, Maxima and minima, Costs Pages: 9 (1051 words) Published: November 15, 2013
MGSC 1206 Test 1 Review Solutions
Not covered in this review:
• Find R(p)
• graphing R(p)
• graphing in general, try graphing in Q#1 – R(x) with C(x), also P(x) • Odds
• Optimization Application Problem involving single variable • Building a demand function and then a revenue function
• Definitions e.g. Non-linear and dynamic functions; Decision-making under certainty, uncertainty and risk
Practice Questions:

x2
20
(a) Find an expression for the marginal revenue first using the limit definition and again using rules for differentiation.

1. Revenue function of producing and selling x units of a product is: R(x) = 20 x −

R(x+h) = 20( x + h) −

( x + h) 2
( x 2 + 2 xh + h 2 )
= 20 x + 20h −
20
20

( x 2 + 2 xh + h 2 )
x2
2 xh + h 2
− 20 x +
= 20h −
R(x+h) – R(x) = 20 x + 20h −
20
20
20
R’(x) = lim 20 − 0.1x + h = 20 − 0.1x
h→0

R’(x) = 20 – 0.1x

(b) Find the quantity that maximizes revenue. Verify it is a maximum. What is the maximum revenue?
R’(x) = 20 – 0.1x = 0
x = 200
R’’(x) = -0.1 < 0 for all x therefore found max
R(200) = 20(200) – (200)2/20 = 2000
(c) The cost function for the product is C(x) = 5x + 500
Form the profit function P(x).

x2
P(x) = R(x) – C(x) = 20 x −
- 5x – 500 = -0.05x2 + 15x -500
20
(d) Find the minimum number of units to break even.

x=

− 15 ± 15 2 − 4(−0.05)(−500) − 15 ± 11.18
=
2(−0.05)
− 0.1

x = 38.2 or x = 261.8
Thus the minimum number of units to break even is 38.2 i.e. 39 units. (e) At what quantity will profit be maximized? Verify. What is the maximum profit? P’(x) = -0.1x + 15 = 0
x = 150
P’’(x) = -0.1 < 0 for all x thus found max
P(150) = -0.05 (150)2 + 15(150) – 500 = 625
2. A box with square top and bottom (with x being the length of the sides of the squares,
and y being the height of the box) is to be made to contain 360 cubic centimetres. Material for the top and bottom costs $ 0.50 per square centimetre and material for the
sides costs $ 0.30 per square centimetre.
(a) Determine a function describing the total cost of the box in terms of its dimensions x and y.
C(x, y) = 0.50x2 + 0.50x2 + (4) 0.30xy = x2 + 1.2xy
(b) Using the volume, rewrite this cost function just in terms of x. V(x, y) = x2y = 360 cm3
C(x) = x2 + 1.2x

360
x

2

y = 360/x2

= x 2 + 432 x −1

(c) Determine all dimensions of the box that minimizes the total cost of material used.
C’(x) = 2x – 432x -2 = 0
x3 = 216
x=6
864
C’’(x) = 2 +
> 0 for x > 0
x3
y = 360/62 = 10
cost at min = 62 + 432/6 = 36 + 72 = $108

3. The following table summarizes the results of a survey of 500 students and faculty at a particular university and their views on the newly implemented screening system which determines the level of financial support provided by the government. They are classified in two ways:

• by generation: “X” – generation X (18-30 years old), “T” – post boomers (31-42), “B” – Boomers (43-60), “W” – Wartime (>60)
• by level of agreement: “H” – high level of agreement, “M” – medium, “L” – low to none.

Level of
Agreement

W

Generation
B
T
X

Total

L

9

16

74

93

192

M

12

34

109

66

221

H

14

38

22

13

87

35

88

205

172

500

Total

Assuming that these survey results are representative of Nova Scotians, determine the following (using probability notation in all answers):
(a) The probability that a randomly selected Nova Scotian is a Gen-X. 172
= 0.344
500

P( X ) =

There is a 34.4% chance that a randomly selected Nova Scotian is a Gen-X.

(b) The probability that a randomly selected Nova Scotian is not a post-boomer. P (T ) = 1 − P (T ) = 1 −

205
= 1 − 0.41 = 0.59
500

The probability is 59% that a randomly selected Nova Scotian is not a post-boomer.

(c) The probability that a randomly selected Nova Scotian has a high level of agreement and is a post-boomer or boomer.
P[H ∩ (T ∪ B )] =

38 + 22
= 0.12
500

The probability is...
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