Question

Steve's utility function is U = BC, where B = beer cans per week and C = pack of cigarettes per week. As a result, his marginal rate of substitution is MRS = -B/C, where beer is on the vertical axis and cigarettes are on the horizontal axis. Steve's income is $120, the price of a can of beer is $2 and that of a pack of cigarettes is $1. [In answering the following, use graphs and math.]
A. How many cans of beer and packs of cigarette does Steve consume? (12)
B. Due to a new tax, the price to Steve of a can of beer rises to $3. Now how much beer and how many packs of cigarettes does Steve consume? (13)

Answer 1

Question #63692 - Economics - Microeconomics | Completed

Question

Steve's utility function is $U = BC$, where $B =$ beer cans per week and $C =$ pack of cigarettes per week. As a result, his marginal rate of substitution is $MRS = -B/C$, where beer is on the vertical axis and cigarettes are on the horizontal axis. Steve's income is $120$, the price of a can of beer is $2 and that of a pack of cigarettes is $1. [In answering the following, use graphs and math.]

A. How many cans of beer and packs of cigarette does Steve consume? (12)

B. Due to a new tax, the price to Steve of a can of beer rises to $3. Now how much beer and how many packs of cigarettes does Steve consume? (13)

Answer

A) Optimal Steve's consumption is in the point $MU_B / P_B = MU_C / P_C$,

Steve's budget equation: $I = P_{B} * B + P_{C} * C \Rightarrow 2B + C = 120$,

where $MU_B = \frac{dU}{dB} = C$, $MU_C = \frac{dU}{dC} = B$

$$\begin{array}{l} M U _ {B} / P _ {B} = M U _ {C} / P _ {C} \Rightarrow C / 2 = B, \text{ put } "B = C / 2" \text{ into } 2B + C = 120 \Rightarrow 2 * (C / 2) + C = 120 \Rightarrow 2C = 120 \Rightarrow \\ \Rightarrow C = 60, B = C / 2 = 30 \\ \end{array}$$

On the graph:

B) Steve's budget equation: $I = P_{B} * B + P_{C} * C \Rightarrow 3B + C = 120$ ,

where $\mathsf{MU}_{\mathsf{B}} = \mathsf{dU} / \mathsf{dB} = \mathsf{C}$ , $\mathsf{MU}_{\mathsf{C}} = \mathsf{dU} / \mathsf{dC} = \mathsf{B}$

$\mathsf{MU}_{\mathsf{B}} / \mathsf{P}_{\mathsf{B}} = \mathsf{MU}_{\mathsf{C}} / \mathsf{P}_{\mathsf{C}} \Rightarrow \mathsf{C} / 3 = \mathsf{B}$ , put “ $\mathsf{B} = \mathsf{C} / 3$ ” into $3\mathsf{B} + \mathsf{C} = 120 \Rightarrow 3^{*}(\mathsf{C} / 3) + \mathsf{C} = 120 \Rightarrow 2\mathsf{C} = 120 \Rightarrow$

$\Rightarrow$ C=60, B=C/3=20

On the graph:

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