<p>I’ll use <span class="math-container">$x:= x_1$</span>, <span class="math-container">$y:= x_2$</span> as they're easier and less confusing to write.</p>
<p>Without the car, you’d solve</p>
<p><span class="math-container">$\max 4 x^{0.5} + 2 y^{0.5}$</span></p>
<p>subject to</p>
<p><span class="math-container">$4 x + 2 y = 20$</span></p>
<p>The optimality condition is <span class="math-container">$MRS = $</span> Relative prices. (This is a shortcut to forming the Lagrangian)</p>
<p><span class="math-container">$\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 $</span></p>
<p>Plugging this in the budget constraint,</p>
<p><span class="math-container">$6x = 20 \implies x^\star = \frac{10}{3} \implies y^\star = \frac{10}{3}$</span></p>
<p>Plugging these optimal consumptions in the utility function,</p>
<p><span class="math-container">$U^\star = 4 (\frac{10}{3})^{0.5} + 2 (\frac{10}{3})^{0.5} = 6 (\frac{10}{3})^{0.5}$</span></p>
<p>On the other hand, with the car you’d solve</p>
<p><span class="math-container">$\max 4 x^{0.5} + 2 y^{0.5}$</span></p>
<p>subject to</p>
<p><span class="math-container">$x + 2 y = 20 - p_{car}$</span></p>
<p>Starting from the optimality condition,</p>
<p><span class="math-container">$\frac{\frac{\partial U}{\partial x}}{\frac{\partial U}{\partial y}} = \frac{p_x}{p_y} \implies \frac{2 x^{-0.5}}{y^{-0.5}} = \frac{1}{2} \implies 4 x^{-0.5} = y^{-0.5} \implies \frac{1}{16} x = y \implies x = 16 y$</span></p>
<p>Plugging this in the budget constraint,</p>
<p><span class="math-container">$18 y = 20 - p_{car} \implies y^\star = \frac{1}{18} (20 - p_{car}) \implies x^\star = \frac{8}{9} (20-p_{car})$</span></p>
<p>Plugging these optimal consumptions in the utility function,</p>
<p><span class="math-container">$U^\star = 4 (\frac{8}{9} (20 - p_{car}))^{0.5} + 2 (\frac{1}{18} (20 - p_{car}))^{0.5} = \frac{8}{3} \sqrt{2} (20 - p_{car})^{0.5} + \frac{2}{3 \sqrt{2}} (20-p_{car})^{0.5} = \frac{8}{3} \sqrt{2} (20 - p_{car})^{0.5} + \frac{1}{3} \sqrt{2} (20 - p_{car})^{0.5} = 3 \sqrt{2} (20 - p_{car})^{0.5}$</span></p>
<p>Now we compare the utility levels for both alternatives and solve for <span class="math-container">$p_{car}$</span>.</p>
<p>In order for the car to be worth it, we need that</p>
<p><span class="math-container">$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} $</span></p>
<p>From here you can conclude that the maximum price you'd be willing to pay for the car is <span class="math-container">$\frac{40}{3}$</span>.</p>

CC BY-SA 4.0
2023-03-15T14:26:47.317Z
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
Fjeeds Arcade
CC BY-SA 4.0
2023-03-15T11:52:22.183Z
2023-03-15T12:00:40.627Z
2023-03-15T12:00:40.627Z
Nicolas Torres
<p>How do i use reservation price to determine when the car ride is the best option?</p>
<p>Assume we have utility function <span class="math-container">$$U(x_1,x_2)=4x_1^{0.5}+2x_2^{0.5}$$</span>
and we have income <span class="math-container">$m=20$</span>. If you stay in Amsterdam for the weekend you can buy 2 goods for the prices <span class="math-container">$p_1=4$</span> and <span class="math-container">$p_2=2$</span>. Alternatively, you can rent a car and drive to Flensburg and purchase the same items for prices <span class="math-container">$p_1=1$</span> and <span class="math-container">$p_2=2$</span>. (Assume that the car ride doesn't add towards utility).</p>
<p>The question is: What is the maximum price you are willing to pay to rent the car? (fuel expenses included)</p>

<microeconomics>
reservation price
Fjeeds Arcade
# reservation price

(Asked by: username_0 on 2023-03-13)

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)

**Tags:** microeconomics


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## 1 answers

### Answer by username_1 on 2023-03-15. Score: 2 | Accepted answer

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 $\frac{40}{3}$.

> **Comments:**
> **username_0**: 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  

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