reservation price

Question

How do i use reservation price to determine when the car ride is the best option?

Assume we have utility function $$U(x_1,x_2)=4x_1^{0.5}+2x_2^{0.5}$$ and we have income $m=20$. If you stay in Amsterdam for the weekend you can buy 2 goods for the prices $p_1=4$ and $p_2=2$. Alternatively, you can rent a car and drive to Flensburg and purchase the same items for prices $p_1=1$ and $p_2=2$. (Assume that the car ride doesn't add towards utility).

The question is: What is the maximum price you are willing to pay to rent the car? (fuel expenses included)

Answer 1

I’ll use $x:= x_1$, $y:= x_2$ as they're easier and less confusing to write.

Without the car, you’d solve

$\max 4 x^{0.5} + 2 y^{0.5}$

subject to

$4 x + 2 y = 20$

The optimality condition is $MRS = $ Relative prices. (This is a shortcut to forming the Lagrangian)

$\frac{\frac{\partial U}{\partial x}}{\frac{\partial U}{\partial y}} = \frac{p_x}{p_y} \implies \frac{2 x^{-0.5}}{y^{-0.5}} = 2 \implies x^{-0.5} = y^{-0.5} \implies x = y $

Plugging this in the budget constraint,

$6x = 20 \implies x^\star = \frac{10}{3} \implies y^\star = \frac{10}{3}$

Plugging these optimal consumptions in the utility function,

$U^\star = 4 (\frac{10}{3})^{0.5} + 2 (\frac{10}{3})^{0.5} = 6 (\frac{10}{3})^{0.5}$

On the other hand, with the car you’d solve

$\max 4 x^{0.5} + 2 y^{0.5}$

subject to

$x + 2 y = 20 - p_{car}$

Starting from the optimality condition,

$\frac{\frac{\partial U}{\partial x}}{\frac{\partial U}{\partial y}} = \frac{p_x}{p_y} \implies \frac{2 x^{-0.5}}{y^{-0.5}} = \frac{1}{2} \implies 4 x^{-0.5} = y^{-0.5} \implies \frac{1}{16} x = y \implies x = 16 y$

Plugging this in the budget constraint,

$18 y = 20 - p_{car} \implies y^\star = \frac{1}{18} (20 - p_{car}) \implies x^\star = \frac{8}{9} (20-p_{car})$

Plugging these optimal consumptions in the utility function,

$U^\star = 4 (\frac{8}{9} (20 - p_{car}))^{0.5} + 2 (\frac{1}{18} (20 - p_{car}))^{0.5} = \frac{8}{3} \sqrt{2} (20 - p_{car})^{0.5} + \frac{2}{3 \sqrt{2}} (20-p_{car})^{0.5} = \frac{8}{3} \sqrt{2} (20 - p_{car})^{0.5} + \frac{1}{3} \sqrt{2} (20 - p_{car})^{0.5} = 3 \sqrt{2} (20 - p_{car})^{0.5}$

Now we compare the utility levels for both alternatives and solve for $p_{car}$.

In order for the car to be worth it, we need that

$u_{car}^\star \geq u_{no car}^\star \implies 3 \sqrt{2} (20 - p_{car})^{0.5} \geq 6 (\frac{10}{3})^{0.5} \implies 18 (20 - p_{car}) \geq 36 \cdot \frac{10}{3} \implies 20 - p_{car} \geq \frac{20}{3} \implies \frac{40}{3} \geq p_{car} \implies p_{car} \leq \frac{40}{3} $

From here you can conclude that the maximum price you'd be willing to pay for the car is $\frac{40}{3}$.