# MGSC: Cube Centimeters Test

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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