(a) (b) (c) 1/3 1/3 Derek has utility for good 1 and good 2 given by u = x₁x₂. What is Derek's Marginal Rate of Substitution, d2 ? dx₁ When ₁2 and 2 = 8, how much of good 2 is Derek willing to give up to get 1 extra unit of good 1? = 8 and ₂ When 1 get 1 extra unit of good 1? = 1, how much of good 2 is Derek willing to give up to

Microeconomics: Principles & Policy
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ISBN:9781337794992
Author:William J. Baumol, Alan S. Blinder, John L. Solow
Publisher:William J. Baumol, Alan S. Blinder, John L. Solow
Chapter5: Consumer Choice: Individual And Market Demand
Section: Chapter Questions
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can you explain how they got the answers here

10.
Derek has utility for good 1 and good 2 given by u = x1/³x/³.
(d)
(e)
What is Derek's Marginal Rate of Substitution, dr2?
When ₁ = 2 and x2 = 8, how much of good 2 is Derek willing to give up to
get 1 extra unit of good 1?
(c)
When x₁ = 8 and x₂ = 1, how much of good 2 is Derek willing to give up to
get 1 extra unit of good 1?
Explain why the two rates in parts (b) and (c) are different.
How are your answers to the above questions different for Derek's friend Joe,
who has utility given by v = x²/³x2/³?
Solution: a) dx2
X2
I1
dx2
X2
⇒ dx₂
dx₁
X1
-4dx₁ ⇒ dx₂
-4(1) ⇒ 4 units. c)
-(1/4) dx₁ ⇒ dx2
−(1/4)(1) ⇒ 0.25 units. d) When he has a lot of x2 and a little of x₁ in part
(b), his marginal utility of x2 is relatively low and his marginal utility of £₁
is relatively high. Therefore, he is willing to give up a lot of x2 to get an
additional unit of x₁. The opposite is true in part (c). e) They would not
change; Joe's utility is a monotonic transformation of Derek's.
dx2
dx1
=
X2
I1
→dx₂
-
=
=
Transcribed Image Text:10. Derek has utility for good 1 and good 2 given by u = x1/³x/³. (d) (e) What is Derek's Marginal Rate of Substitution, dr2? When ₁ = 2 and x2 = 8, how much of good 2 is Derek willing to give up to get 1 extra unit of good 1? (c) When x₁ = 8 and x₂ = 1, how much of good 2 is Derek willing to give up to get 1 extra unit of good 1? Explain why the two rates in parts (b) and (c) are different. How are your answers to the above questions different for Derek's friend Joe, who has utility given by v = x²/³x2/³? Solution: a) dx2 X2 I1 dx2 X2 ⇒ dx₂ dx₁ X1 -4dx₁ ⇒ dx₂ -4(1) ⇒ 4 units. c) -(1/4) dx₁ ⇒ dx2 −(1/4)(1) ⇒ 0.25 units. d) When he has a lot of x2 and a little of x₁ in part (b), his marginal utility of x2 is relatively low and his marginal utility of £₁ is relatively high. Therefore, he is willing to give up a lot of x2 to get an additional unit of x₁. The opposite is true in part (c). e) They would not change; Joe's utility is a monotonic transformation of Derek's. dx2 dx1 = X2 I1 →dx₂ - = =
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Follow-up Question

can you now solve part d and e

10.
(a)
(c)
1/3 1/3
Derek has utility for good 1 and good 2 given by u = x₁x₂
dx2?
What is Derek's Marginal Rate of Substitution,
When 1 =
2 and 2
8, how much of good 2 is Derek willing to give up to
get 1 extra unit of good 1?
=
When 1 = 8 and ₂ =
get 1 extra unit of good 1?
Explain why the two rates in parts (b) and (c) are different.
How are your answers to the above questions different for Derek's friend Joe,
who has utility given by v = x²/³x2/³?
Solution:
dx2
dz₁
1, how much of good 2 is Derek willing to give up to
dx2
dx₁
x2
I1
−4(1) ⇒ 4 units. c) da
⇒ dx₂
-4dx₁ ⇒ dx2
-(1/4)dx₁ ⇒ dx₂
-(1/4)(1) ⇒ 0.25 units. d) When he has a lot of x2 and a little of x₁ in part
(b), his marginal utility of x2 is relatively low and his marginal utility of x₁
is relatively high. Therefore, he is willing to give up a lot of 2 to get an
additional unit of ₁. The opposite is true in part (c). e) They would not
change; Joe's utility is a monotonic transformation of Derek's.
X2
x2
I1
=
⇒ dx₂
-
=
=
=
Transcribed Image Text:10. (a) (c) 1/3 1/3 Derek has utility for good 1 and good 2 given by u = x₁x₂ dx2? What is Derek's Marginal Rate of Substitution, When 1 = 2 and 2 8, how much of good 2 is Derek willing to give up to get 1 extra unit of good 1? = When 1 = 8 and ₂ = get 1 extra unit of good 1? Explain why the two rates in parts (b) and (c) are different. How are your answers to the above questions different for Derek's friend Joe, who has utility given by v = x²/³x2/³? Solution: dx2 dz₁ 1, how much of good 2 is Derek willing to give up to dx2 dx₁ x2 I1 −4(1) ⇒ 4 units. c) da ⇒ dx₂ -4dx₁ ⇒ dx2 -(1/4)dx₁ ⇒ dx₂ -(1/4)(1) ⇒ 0.25 units. d) When he has a lot of x2 and a little of x₁ in part (b), his marginal utility of x2 is relatively low and his marginal utility of x₁ is relatively high. Therefore, he is willing to give up a lot of 2 to get an additional unit of ₁. The opposite is true in part (c). e) They would not change; Joe's utility is a monotonic transformation of Derek's. X2 x2 I1 = ⇒ dx₂ - = = =
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