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The utility function seems innocent: $u(x_1,x_2)=(x_1^2+x_1)x_2$. I want to find the indirect utility function and the expenditure function, but I've encountered some problems. Here's what I've got: $$MU_1=2x_1x_2+x_2,MU_2=x_1^2+x_1$$ $$\frac{MU_1}{MU_2}=\frac{p_1}{p_2}\implies \frac{2x_1x_2+x_2}{x_1^2+x_1}=\frac{p_1}{p_2}$$ $$\implies x_1^2p_1+x_1(p_1-2x_2p_2)-x_2p_2=0$$ $$\implies x_1=\frac{2x_2p_2-p_1+\sqrt{p_1^2+4x_2^2p_2^2}}{2p_1}$$ $$x_2=\frac{x_1^2p_1+x_1p_1}{2x_1p_2+p_2}$$ Substitute back into the budget constraint, I got $$W=\frac{2x_2p_2-p_1+\sqrt{p_1^2+4x_2^2p_2^2}}{2}+p_2x_2$$ but I have no idea how to solve for $x_2$. Maybe I'm going in the wrong direction. Any ideas are greatly appreciated.

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1 Answer 1

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Rather than solving the tangency condition for $x_1$ and substituting into the budget constraint, it is much easier to solve the tangency condition for $x_2$ and substitute into the budget constraint.

As you have found, the tangency condition requires that

$$x_2=\frac{p_1x_1(x_1+1)}{p_2(2x_1+1)}$$

Substituting into the budget constraint gives

$$p_1x_1+p_2\frac{p_1x_1(x_1+1)}{p_2(2x_1+1)}=W.$$

Multiplying through by $(2x_1+1)$ gives

$$p_1x_1(2x_1+1)+p_1x_1(x_1+1)=(2x_1+1)W$$

or

$$p_1x_1(3x_1+2)=(2x_1+1)W$$

Now just rearrange to get a quadratic equation in $x_1$ and solve using the quadratic formula.


Note that you should also argue why it is not optimal for $x_1$ or $x_2$ to be zero (i.e. why a corner solution is not optimal).

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  • $\begingroup$ But how to find the expenditure function? $u=(x_1^2+x_1)x_2=(x_1^2+x_1)\frac{x_1p_1(x_1+1)}{p_2(2x_1+1)}\implies p_1 x_1^4+2p_1x_1^3+p_1x_1^2-2up_2x_1=p_2u$. How do I solve for $x_1$? $\endgroup$ Sep 8, 2023 at 0:50
  • $\begingroup$ Once you have found the solution $x_1^*(p,W)$, $x_2^*(p,W)$ to the utility maximization problem, you find the indirect utility function $v$, which is given by $v(p,W)=u(x_1^*(p,W),x_2^*(p,W))$, by substituting the solution $(x_1^*,x_2^*)$ into $u$. Then, to find the expenditure function $e$, use the equation $v(p,e(p,\bar{u}))=\bar{u}$ (which amounts to solving $v(p,e)=\bar{u}$ for $e$). $\endgroup$
    – smcc
    Sep 8, 2023 at 4:35

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