# Derive the Hicks demand function for $U(x_1,x_2) = x_1^{1/2}x_2^{1/3}$

Consider the utility function $$U(x_1,x_2) = x_1^{1/2}x_2^{1/3}$$ and the budget line $$p_1x_1+p_2x_2 = m$$. Then I have to find the Hicks demand function.

I know that to do this I have to solve the following expenditure minimization problem $$\min_{x_1,x_2} p_1x_1+p_2x_2$$ subject to $$x_1^{1/2}x_2^{1/3} = U$$ for some utility level $$U$$. I have already found the Marshallian demand functions to be $$x_1^* = \frac{3m}{5p_1}$$ and $$x^* = \frac{2m}{5p_2}$$.

I thought maybe we could use the well-known Cobb-Douglas demand functions but I don't think we can as $$1/2 + 1/3 = 5/6 \neq 1$$.

I then tried to use Lagrange and got that

By Lagrange we have that $$L = p_1x_1+p_2x_2 - \lambda \left( x_1^{1/2}x_2^{1/3}-U \right)$$ Thus the three FOC will be \begin{align} \frac{\partial L}{\partial x_1}: p_1- \lambda \left( \frac{1}{2} x_1^{-1/2}x_2^{1/3} \right) \\ \frac{\partial L}{\partial x_2}: p_2 - \lambda \left( \frac{1}{3} x_1^{1/2}x_2^{-2/3} \right) \\ \frac{\partial L}{\partial \lambda}: x_1^{1/2}x_2^{1/3} = U \end{align} Dividng the first equation with the second equation one gets $$\frac{3x_2}{2x_1} = \frac{p_1}{p_2}$$ Thus $$x_1 = \frac{3p_2x_2}{2p_1}$$ Inserting this into $$U(x_1,x_2)$$ we have that $$\left(\frac{3p_2x_2}{2p_1} \right)^{1/2} x_2^{1/3}$$ which gives me after some few calculations that $$x_1^* = \frac{3p_2}{2p_1} \left( \frac{U\sqrt{2}\sqrt{p_1}}{ \sqrt{3} \sqrt{p_2}} \right)^{6/5}$$ $$x_2^* = \left( \frac{ U \sqrt{2} \sqrt{p_1}}{ \sqrt{3} \sqrt{p_2} } \right)^{1/(5/6)}$$ but this doesn't seem right. Is there an easier solution to this kind of problem?

Let $$f(U)=U^{6/5}$$. This is a positive monotone transformation of $$U$$ on $$\mathbb{R}_0^+$$. So the preferences represented by $$U$$ are also represented by $$V(x_1,x_2):=f(U(x_1,x_2))=x_1^{3/5}x_2^{2/5}$$. The utility function $$V$$ has Cobb-Douglas form and you can use the formula for the Hicksian demand for Cobb-Douglas utilities: $$x_1^*=\left(\frac{3p_2}{2p_1}\right)^{2/5}V,\quad x_2^*=\left(\frac{2p_1}{3p_2}\right)^{3/5}V.$$ Substituting for $$V$$ you get $$x_1^*=\left(\frac{3p_2}{2p_1}\right)^{2/5}U^{6/5},\quad x_2^*=\left(\frac{2p_1}{3p_2}\right)^{3/5}U^{6/5},$$ or $$x_1^*=\left[\left(\frac{3p_2}{2p_1}\right)U^3\right]^{2/5},\quad x_2^*=\left[\left(\frac{2p_1}{3p_2}\right)U^2\right]^{3/5}.$$