Is my solution to finding the Hicksian demand correct? Maximize $x_1^{1/2} + x_2^{1/2}$ subejct to the budget constraint

Question

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)

Answer 1

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

$$\frac{p_1^{-2}}{(p_1^{-1}+p_2^{-1} )^2}u^2=x_1$$

Answer 2

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: $(x_1^h,x_2^h)(p_1,p_2,\mu)=\left(\left(\frac{p_2\mu}{p_1+p_2}\right)^2,\left(\frac{p_1\mu}{p_1+p_2}\right)^2 \right)$