# reservation price

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)

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 $$\frac{40}{3}$$.

• Thank you so much! This made a lot of sense. I thought at first that I could just solve the regular problem, which gave me x = y =10/3 and then say 20 - 2(10/3) - 10/3. But this obviously doesn't account for the chancing prices Mar 15 at 14:26