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?


1 Answer 1


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 $$x_1^*=\left[\left(\frac{3p_2}{2p_1}\right)U^3\right]^{2/5},\quad x_2^*=\left[\left(\frac{2p_1}{3p_2}\right)U^2\right]^{3/5}.$$


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