# Utility function and consumer's optimal demand (edited)

Derive the consumer's optimal demand of x and y, respectively, when the budget constraint is $$px + qy = m$$ and the utility function is`u = $$x^p + y^p$$

here's my attempt but i'm lost on how to solve for $$x$$. Can someone please help with simple and easy to follow instructions?

I'm lost after that. Please help me with steps. The answer for $$x=$$ $$((p/q))^((1/(p-1))$$ * $$y$$.

that's supposed to be power of $$^(1/p-1)$$ btw. i can solve the rest but idk where the power of $$^(1/p-1)$$ came from.

• Hi! Unless there is a typo here, this is not a Cobb-Douglas utility function. Also, are you sure the powers in the function are exactly equal to the price of good $x$, that is, $p$? If yes, are there any restrictions on this parameter? Please check your question for typo's and edit where necessary. Commented Oct 26, 2021 at 6:23
• @Giskard Hi, I fixed it. Hopefully it's better now. Commented Oct 27, 2021 at 19:58
• Are you sure the powers in the function are exactly equal to the price of good $x$, that is, $p$? If yes, are there any restrictions on this parameter? Commented Oct 28, 2021 at 10:00

I think there are some typos in your question. First of all, this is not a Cobb-Douglas function. Second, the exponents seem to be incorrect.

Nevertheless, let us change the utility function to a Cobb-Douglas case (as was meant in your question):

\begin{align} u(x, y) = x^\alpha y^{1-\alpha}. \end{align}

Let us, for the sake of readability, change the prices to $$p_x$$ (instead of $$p$$) and $$p_y$$ (instead of q). The optimization problem is quite simple in this case and reads $$$$\max_{x, y} \mathcal{L} = x^\alpha y^{1-\alpha}- \lambda\left(p_xx + p_yy - m\right).$$$$

The corresponding first-order conditions read \begin{align} \frac{\partial \mathcal{L}}{\partial x} &= \alpha x^{\alpha-1}y^{1-\alpha}- \lambda p_x =0 \\ \frac{\partial \mathcal{L}}{\partial y} &= (1-\alpha)x^{\alpha}y^{-\alpha} - \lambda p_y =0 \end{align}

It follows that \begin{align} y(x) &= \frac{1-\alpha}{\alpha}\frac{p_x}{p_y}x \\ x(y) &= \frac{\alpha}{1-\alpha}\frac{p_y}{p_x}y \\ \end{align}

All you need to do is to use the expression for $$x$$ and put it into your budget constraint to find a solution for $$y$$. Alternatively, you can use the expression for $$y$$ and find a solution for $$x$$. You will get exactly the same results. In the latter case, you will find \begin{align} m = p_xx + p_y\frac{1-\alpha}{\alpha}\frac{p_x}{p_y}x, \end{align} which simplifies to \begin{align} m &= x\left(p_x\left[1+\frac{1-\alpha}{\alpha}\right]\right)\\ m &= x\frac{p_x}{\alpha} \end{align} Your solution for $$x^\star$$ is then given by $$x^\star=\frac{\alpha}{p_x}m$$. Put this expression for $$x^\star$$ into the expression for $$y(x)$$ from above to get $$y^\star = \frac{1-\alpha}{p_y}m$$.

• I've edited the question and fixed the typos, hopefully it's better now! Commented Oct 28, 2021 at 1:45