<p>Here is what I am thinking (and i could have made a mistake since CES algebra is a bit tedious)</p>
<p>First you have the MRS condition from Lagrange which boils down to</p>
<p><span class="math-container">$$p_1 x_2^{\rho-1} = p_2 x_1^{\rho-1}$$</span></p>
<p>from here I always aim to reconstruct the utility function (because it is a constant equal to the max utility or in this instance you know it will satisfy constraint and be u). So to get the utility function to appear I do this</p>
<p><span class="math-container">$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} = p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$</span></p>
<p>I then add <span class="math-container">$p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}$</span> to both sides</p>
<p><span class="math-container">$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} + p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}= p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}+p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$</span></p>
<p>and isolate</p>
<p><span class="math-container">$$p_1^{\frac{\rho}{\rho-1}}( x_2^{\rho} +  x_1^{\rho})= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$</span></p>
<p>utility function appears on LHS and I impose constraint
<span class="math-container">$$p_1^{\frac{\rho}{\rho-1}}u= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$</span></p>
<p>finding</p>
<p><span class="math-container">$$\frac{p_1^{\frac{\rho}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )}u=x_1^{\rho}$$</span></p>
<p>which reduces to</p>
<p><span class="math-container">$$\frac{p_1^{\frac{1}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )^\frac{1}{\rho}}u^\frac{1}{\rho}=x_1$$</span></p>
<p>Inserting value of <span class="math-container">$\rho = 1/2$</span> I get</p>
<p><span class="math-container">$$\frac{p_1^{-2}}{(p_1^{-1}+p_2^{-1} )^2}u^2=x_1$$</span></p>

CC BY-SA 4.0
2020-12-20T21:24:08.420Z
2020-12-20T21:24:08.420Z
Jesper Hybel
<p>You can just solve the following problem to obtain Hicksian Demands directly:
<span class="math-container">$$\begin{aligned}
&amp;&amp;\min_{x_1,x_2\geq0} \quad &amp; p_1x_1+p_2x_2\\
&amp;&amp;\textrm{s.t.} \quad &amp; \sqrt x_1 + \sqrt x_2 \geq \mu\\\\
&amp;&amp;\min_{x_1\geq0} \quad &amp;p_1x_1+p_2(\mu-\sqrt x_1)^2 \\
\end{aligned}$$</span></p>
<p>we get the second problem involving only <span class="math-container">$x_1$</span> by using the substitution <span class="math-container">$\sqrt x_2 =\mu -\sqrt x_1$</span> since the constraint binds at optimum.</p>
<p>Now, let <span class="math-container">$f(x_1)=p_1x_1+p_2(\mu-\sqrt x_1)^2$</span> defined for <span class="math-container">$x_1\geq 0$</span>.
Let us see whether <span class="math-container">$f(x_1)$</span> is convex or concave over its entire domain.</p>
<p><span class="math-container">$$\begin{eqnarray}
f'(x_1)=p_1+p_2-\frac{p_2\mu}{\sqrt x_1}\\
f''(x_1)=\frac{p_2 \mu}{2x_1^{\frac{3}{2}}}&gt;0 &amp; \quad \text{for }x_1\geq 0
\end{eqnarray}$$</span></p>
<p>since <span class="math-container">$f(x_1)$</span> is convex over its entire domain if there exists a stationary point <span class="math-container">$x_1^*\geq 0$</span> then it's the solution to our problem.
Stationary points can be obtained by:
<span class="math-container">$f(x_1)\overset{set}{\equiv}0$</span> yields <span class="math-container">$x_1=\left(\frac{p_2\mu}{p_1+p_2}\right)^2\geq 0$</span></p>
<p>Therefore, hicksian demand functions are: <span class="math-container">$(x_1^h,x_2^h)(p_1,p_2,\mu)=\left(\left(\frac{p_2\mu}{p_1+p_2}\right)^2,\left(\frac{p_1\mu}{p_1+p_2}\right)^2 \right)$</span></p>

CC BY-SA 4.0
2023-04-25T15:11:35.100Z
2023-06-04T16:12:14.850Z
2023-06-04T16:12:14.850Z
mynameparv
<p>Maximize <span class="math-container">$x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}$</span> subejct to the budget constraint <span class="math-container">$p_1x_1+p_2x_2=m$</span></p>
<p>Setting up the Lagrange and finding the first-order conditions:</p>
<p><span class="math-container">$L(x_1, x_2, \lambda)=x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}+\lambda(p_1x_1+p_2x_2-m)$</span></p>
<p><span class="math-container">$\frac{\partial L}{\partial x_1}=\frac{1}{2}x_1^\frac{-1}{2}+\lambda p_1=0$</span></p>
<p><span class="math-container">$\frac{\partial L}{\partial x_2}=\frac{1}{2}x_2^\frac{-1}{2}+\lambda p_2=0$</span></p>
<p>Equating these,</p>
<p><span class="math-container">$2p_2x_1^\frac{-1}{2}=2p_1x_2^\frac{-1}{2}$</span></p>
<p>Solving for <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>, we get</p>
<p><span class="math-container">$x_1=\frac{p_2^2x_2}{p_1^2}$</span></p>
<p><span class="math-container">$x_2=\frac{p_1^2x_1}{p_2^2}$</span></p>
<p>Substituting these into the budget constraint and solving to find the Hicksian demand (<span class="math-container">$x^*(p,m)$</span>),</p>
<p><span class="math-container">$p_1x_1+p_2x_2=m$</span></p>
<p><span class="math-container">$\frac{p_2^2x_2}{p_1}+p_2x_2=m$</span></p>
<p>Solving for <span class="math-container">$x_2$</span>,</p>
<p><span class="math-container">$x_2(\frac{p_2^2}{p_1}+p_2)=m$</span></p>
<p><span class="math-container">$x_2^*=\frac{m}{p_2}.\frac{p_2}{p_1+p_2}=\frac{m}{p_1+p_2}$</span></p>
<p>We do the same thing for <span class="math-container">$x_1$</span> and we get</p>
<p><span class="math-container">$x_1^*=\frac{m}{p_1+p_2}$</span></p>
<p>We now have the Hickdian demands, substitute these into the objective function to get the indirect utility function and by duality, we know, <span class="math-container">$V(p,E(p,u))=u$</span>,</p>
<p><span class="math-container">$V(p,m)=(\frac{m}{p_1+p_2})^\frac{1}{2}+(\frac{m}{p_1+p_2})^\frac{1}{2}=u$</span></p>
<p>Solving this to find m,</p>
<p><span class="math-container">$\frac{2m^\frac{1}{2}}{p_1^\frac{1}{2}+p_2^\frac12}=u$</span></p>
<p><span class="math-container">$m^\frac{1}{2}=\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})$</span></p>
<p><span class="math-container">$m=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$</span></p>
<p>Finally we also know by duality that <span class="math-container">$E(p,V(p,m))=m$</span>, therefore</p>
<p><span class="math-container">$E=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$</span></p>
<p>To find the Hicksian demand we use Shephards Lemma (take the partial derivative of the expenditure function)</p>
<p><span class="math-container">$\frac{\partial E}{\partial p_i}=2.\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2}).\frac{1}{2}.\frac{u}{2}.p_1^\frac{-1}{2}$</span></p>
<p>Simplifying this</p>
<p><span class="math-container">$h_1^*=\frac{u^2}{4}.p_1^\frac{-1}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})=\frac{u^2}{4}(p_1^\frac{-1}{4}+p_2^\frac{1}{2})$</span></p>
<p>Is this correct? (I've tried to make each of my steps really clear so this is easy to follow)</p>

CC BY-SA 4.0
2020-12-20T20:54:51.727Z
I dont believe your last equation where $p_1$ disappears.
Jesper Hybel
CC BY-SA 4.0
2020-12-20T21:11:14.713Z
@JesperHybel You're right! I've changed it now, thanks
CorporateNationalism
CC BY-SA 4.0
2023-04-25T15:14:17.970Z
Your process is correct, but it's very long. You should just try to solve the expenditure minimization problem, I've solved it in an answer below as an example
mynameparv
CC BY-SA 4.0
2023-04-25T16:45:32.297Z
There is an error in your last equality
Bertrand
<profit-maximization><demand><duality>
Is my solution to finding the Hicksian demand correct? Maximize $x_1^{1/2} + x_2^{1/2}$ subejct to the budget constraint
CorporateNationalism
# Is my solution to finding the Hicksian demand correct? Maximize $x_1^{1/2} + x_2^{1/2}$ subejct to the budget constraint

(Asked by: username_0 on 2020-12-20)

Maximize $x\_1^{\frac{1}{2}} + x\_2^{\frac{1}{2}}$ subejct to the budget constraint $p\_1x\_1+p\_2x\_2=m$

Setting up the Lagrange and finding the first-order conditions:

$L(x\_1, x\_2, \lambda)=x\_1^{\frac{1}{2}} + x\_2^{\frac{1}{2}}+\lambda(p\_1x\_1+p\_2x\_2-m)$

$\frac{\partial L}{\partial x\_1}=\frac{1}{2}x\_1^\frac{-1}{2}+\lambda p\_1=0$

$\frac{\partial L}{\partial x\_2}=\frac{1}{2}x\_2^\frac{-1}{2}+\lambda p\_2=0$

Equating these,

$2p\_2x\_1^\frac{-1}{2}=2p\_1x\_2^\frac{-1}{2}$

Solving for $x\_1$ and $x\_2$, we get

$x\_1=\frac{p\_2^2x\_2}{p\_1^2}$

$x\_2=\frac{p\_1^2x\_1}{p\_2^2}$

Substituting these into the budget constraint and solving to find the Hicksian demand ($x^\*(p,m)$),

$p\_1x\_1+p\_2x\_2=m$

$\frac{p\_2^2x\_2}{p\_1}+p\_2x\_2=m$

Solving for $x\_2$,

$x\_2(\frac{p\_2^2}{p\_1}+p\_2)=m$

$x\_2^\*=\frac{m}{p\_2}.\frac{p\_2}{p\_1+p\_2}=\frac{m}{p\_1+p\_2}$

We do the same thing for $x\_1$ and we get

$x\_1^\*=\frac{m}{p\_1+p\_2}$

We now have the Hickdian demands, substitute these into the objective function to get the indirect utility function and by duality, we know, $V(p,E(p,u))=u$,

$V(p,m)=(\frac{m}{p\_1+p\_2})^\frac{1}{2}+(\frac{m}{p\_1+p\_2})^\frac{1}{2}=u$

Solving this to find m,

$\frac{2m^\frac{1}{2}}{p\_1^\frac{1}{2}+p\_2^\frac12}=u$

$m^\frac{1}{2}=\frac{u}{2}(p\_1^\frac{1}{2}+p\_2^\frac{1}{2})$

$m=[\frac{u}{2}(p\_1^\frac{1}{2}+p\_2^\frac{1}{2})]^2$

Finally we also know by duality that $E(p,V(p,m))=m$, therefore

$E=[\frac{u}{2}(p\_1^\frac{1}{2}+p\_2^\frac{1}{2})]^2$

To find the Hicksian demand we use Shephards Lemma (take the partial derivative of the expenditure function)

$\frac{\partial E}{\partial p\_i}=2.\frac{u}{2}(p\_1^\frac{1}{2}+p\_2^\frac{1}{2}).\frac{1}{2}.\frac{u}{2}.p\_1^\frac{-1}{2}$

Simplifying this

$h\_1^\*=\frac{u^2}{4}.p\_1^\frac{-1}{2}(p\_1^\frac{1}{2}+p\_2^\frac{1}{2})=\frac{u^2}{4}(p\_1^\frac{-1}{4}+p\_2^\frac{1}{2})$

Is this correct? (I've tried to make each of my steps really clear so this is easy to follow)

**Tags:** profit-maximization, demand, duality
> **Comments:**
> **username_1**: I dont believe your last equation where $p_1$ disappears.  
> **username_0**: @JesperHybel You're right! I've changed it now, thanks  
> **username_2**: Your process is correct, but it's very long. You should just try to solve the expenditure minimization problem, I've solved it in an answer below as an example  
> **username_3**: There is an error in your last equality  

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## 2 answers

### Answer by username_1 on 2020-12-20. Score: 2

Here is what I am thinking (and i could have made a mistake since CES algebra is a bit tedious)

First you have the MRS condition from Lagrange which boils down to

$$p\_1 x\_2^{\rho-1} = p\_2 x\_1^{\rho-1}$$

from here I always aim to reconstruct the utility function (because it is a constant equal to the max utility or in this instance you know it will satisfy constraint and be u). So to get the utility function to appear I do this

$$p\_1^{\frac{\rho}{\rho-1}} x\_2^{\rho} = p\_2^{\frac{\rho}{\rho-1}} x\_1^{\rho}$$

I then add $p\_1^{\frac{\rho}{\rho-1}} x\_1^{\rho}$ to both sides

$$p\_1^{\frac{\rho}{\rho-1}} x\_2^{\rho} + p\_1^{\frac{\rho}{\rho-1}} x\_1^{\rho}= p\_1^{\frac{\rho}{\rho-1}} x\_1^{\rho}+p\_2^{\frac{\rho}{\rho-1}} x\_1^{\rho}$$

and isolate

$$p\_1^{\frac{\rho}{\rho-1}}( x\_2^{\rho} + x\_1^{\rho})= (p\_1^{\frac{\rho}{\rho-1}}+p\_2^{\frac{\rho}{\rho-1}} )x\_1^{\rho}$$

utility function appears on LHS and I impose constraint
$$p\_1^{\frac{\rho}{\rho-1}}u= (p\_1^{\frac{\rho}{\rho-1}}+p\_2^{\frac{\rho}{\rho-1}} )x\_1^{\rho}$$

finding

$$\frac{p\_1^{\frac{\rho}{\rho-1}}}{(p\_1^{\frac{\rho}{\rho-1}}+p\_2^{\frac{\rho}{\rho-1}} )}u=x\_1^{\rho}$$

which reduces to

$$\frac{p\_1^{\frac{1}{\rho-1}}}{(p\_1^{\frac{\rho}{\rho-1}}+p\_2^{\frac{\rho}{\rho-1}} )^\frac{1}{\rho}}u^\frac{1}{\rho}=x\_1$$

Inserting value of $\rho = 1/2$ I get

$$\frac{p\_1^{-2}}{(p\_1^{-1}+p\_2^{-1} )^2}u^2=x\_1$$


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### Answer by username_2 on 2023-04-25. Score: 0

You can just solve the following problem to obtain Hicksian Demands directly:
$$\begin{aligned}
&&\min\_{x\_1,x\_2\geq0} \quad & p\_1x\_1+p\_2x\_2\\
&&\textrm{s.t.} \quad & \sqrt x\_1 + \sqrt x\_2 \geq \mu\\\\
&&\min\_{x\_1\geq0} \quad &p\_1x\_1+p\_2(\mu-\sqrt x\_1)^2 \\
\end{aligned}$$

we get the second problem involving only $x\_1$ by using the substitution $\sqrt x\_2 =\mu -\sqrt x\_1$ since the constraint binds at optimum.

Now, let $f(x\_1)=p\_1x\_1+p\_2(\mu-\sqrt x\_1)^2$ defined for $x\_1\geq 0$.
Let us see whether $f(x\_1)$ is convex or concave over its entire domain.

$$\begin{eqnarray}
f'(x\_1)=p\_1+p\_2-\frac{p\_2\mu}{\sqrt x\_1}\\
f''(x\_1)=\frac{p\_2 \mu}{2x\_1^{\frac{3}{2}}}>0 & \quad \text{for }x\_1\geq 0
\end{eqnarray}$$

since $f(x\_1)$ is convex over its entire domain if there exists a stationary point $x\_1^\*\geq 0$ then it's the solution to our problem.
Stationary points can be obtained by:
$f(x\_1)\overset{set}{\equiv}0$ yields $x\_1=\left(\frac{p\_2\mu}{p\_1+p\_2}\right)^2\geq 0$

Therefore, hicksian demand functions are: $(x\_1^h,x\_2^h)(p\_1,p\_2,\mu)=\left(\left(\frac{p\_2\mu}{p\_1+p\_2}\right)^2,\left(\frac{p\_1\mu}{p\_1+p\_2}\right)^2 \right)$


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