<p>Hint for Question 1.</p>
<p>Taking the partial derivative of the Lagrangian with respect to <span class="math-container">$x$</span> and setting it equal to zero you should have obtained:</p>
<p><span class="math-container">$$\frac{x^{-1/2}y^{1/2}}{2}-\lambda p_x=0$$</span></p>
<p>Then you can substitute the optimal values of <span class="math-container">$x$</span> and <span class="math-container">$y$</span> and solve for <span class="math-container">$\lambda$</span>.  Can you take it from there?</p>

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2022-09-25T10:11:53.310Z
2022-09-25T10:11:53.310Z
Adam Bailey
<p>Background: Suppose a consumer has the following utility function <span class="math-container">$u(x,y)=\sqrt{xy}$</span>, then the Lagrangian equation is <span class="math-container">$\mathcal{L}(x,y)=\sqrt{xy}+\lambda(I-xp_x-yp_y)$</span>. Then the optimal bundle is <span class="math-container">$(x,y)^*=(\frac{I}{2p_x},\frac{I}{2p_y})$</span> and <span class="math-container">$\lambda^*=\frac{1}{2\sqrt{p_xp_y}}$</span>.</p>
<p>Question 1): For <span class="math-container">$x^*,y^*$</span>,I know how to get those,i.e. by solving <span class="math-container">$\frac{\frac{\partial{u_x}}{\partial x}}{\frac{\partial{u_y}}{\partial y}}$</span> and plugging into income equation. I am not sure how to calculate to get <span class="math-container">$\lambda^*$</span>.</p>
<p>Please let me know what context to add to make questions clearer, thanks.</p>

CC BY-SA 4.0
2022-09-25T07:24:10.347Z
1. "$\lambda^*=\frac{1}{2\sqrt{p_xp_y}}$" ,,, " I  am not sure how to get $\lambda$"  ?
Giskard
CC BY-SA 4.0
2022-09-25T07:26:53.963Z
2. I am not sure I understand your question here either. It would be surprising if this formula for $(x,y)*$ would work for every single utility maximization problem; in this sense it is not indeed not *general*. You can indeed verify this by solving the problem for any utility function representing different preferences.
Giskard
CC BY-SA 4.0
2022-09-25T20:23:21.113Z
@Giskard It was on lecture notes, it wrote that the optimal lagrange multiplier was $\lambda^*$. I deleted Question 2, since I figured out.
Twilight
<microeconomics><lagrangian>
Questions about Lagrangian and consumer's problem
Twilight
# Questions about Lagrangian and consumer's problem

(Asked by: username_0 on 2022-09-24)

Background: Suppose a consumer has the following utility function $u(x,y)=\sqrt{xy}$, then the Lagrangian equation is $\mathcal{L}(x,y)=\sqrt{xy}+\lambda(I-xp\_x-yp\_y)$. Then the optimal bundle is $(x,y)^\*=(\frac{I}{2p\_x},\frac{I}{2p\_y})$ and $\lambda^\*=\frac{1}{2\sqrt{p\_xp\_y}}$.

Question 1): For $x^\*,y^\*$,I know how to get those,i.e. by solving $\frac{\frac{\partial{u\_x}}{\partial x}}{\frac{\partial{u\_y}}{\partial y}}$ and plugging into income equation. I am not sure how to calculate to get $\lambda^\*$.

Please let me know what context to add to make questions clearer, thanks.

**Tags:** microeconomics, lagrangian
> **Comments:**
> **username_1**: 1. "$\lambda^*=\frac{1}{2\sqrt{p_xp_y}}$" ,,, " I  am not sure how to get $\lambda$"  ?  
> **username_1**: 2. I am not sure I understand your question here either. It would be surprising if this formula for $(x,y)*$ would work for every single utility maximization problem; in this sense it is not indeed not *general*. You can indeed verify this by solving the problem for any utility function representing different preferences.  
> **username_0**: @Giskard It was on lecture notes, it wrote that the optimal lagrange multiplier was $\lambda^*$. I deleted Question 2, since I figured out.  

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## 1 answers

### Answer by username_2 on 2022-09-25. Score: 2 | Accepted answer

Hint for Question 1.

Taking the partial derivative of the Lagrangian with respect to $x$ and setting it equal to zero you should have obtained:

$$\frac{x^{-1/2}y^{1/2}}{2}-\lambda p\_x=0$$

Then you can substitute the optimal values of $x$ and $y$ and solve for $\lambda$. Can you take it from there?


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