MGSC: Cube Centimeters Test
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
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