<p>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.</p>
<p>Nevertheless, let us change the utility function to a Cobb-Douglas case (as was meant in your question):</p>
<p><span class="math-container">\begin{align}
u(x, y) = x^\alpha y^{1-\alpha}.
\end{align}</span></p>
<p>Let us, for the sake of readability, change the prices to <span class="math-container">$p_x$</span> (instead of <span class="math-container">$p$</span>) and <span class="math-container">$p_y$</span> (instead of q). The optimization problem is quite simple in this case and reads
<span class="math-container">\begin{equation}
\max_{x, y} \mathcal{L} = x^\alpha y^{1-\alpha}- \lambda\left(p_xx + p_yy - m\right).
\end{equation}</span></p>
<p>The corresponding first-order conditions read
<span class="math-container">\begin{align}
\frac{\partial \mathcal{L}}{\partial x} &amp;= \alpha x^{\alpha-1}y^{1-\alpha}- \lambda p_x =0 \\
\frac{\partial \mathcal{L}}{\partial y} &amp;= (1-\alpha)x^{\alpha}y^{-\alpha} - \lambda p_y =0 
\end{align}</span></p>
<p>It follows that
<span class="math-container">\begin{align}
y(x) &amp;= \frac{1-\alpha}{\alpha}\frac{p_x}{p_y}x \\
x(y) &amp;= \frac{\alpha}{1-\alpha}\frac{p_y}{p_x}y \\
\end{align}</span></p>
<p>All you need to do is to use the expression for <span class="math-container">$x$</span> and put it into your budget constraint to find a solution for <span class="math-container">$y$</span>. Alternatively, you can use the expression for <span class="math-container">$y$</span> and find a solution for <span class="math-container">$x$</span>. You will get exactly the same results. In the latter case, you will find
<span class="math-container">\begin{align}
m = p_xx + p_y\frac{1-\alpha}{\alpha}\frac{p_x}{p_y}x,
\end{align}</span>
which simplifies to
<span class="math-container">\begin{align}
m &amp;= x\left(p_x\left[1+\frac{1-\alpha}{\alpha}\right]\right)\\
m &amp;= x\frac{p_x}{\alpha}
\end{align}</span>
Your solution for <span class="math-container">$x^\star$</span> is then given by <span class="math-container">$x^\star=\frac{\alpha}{p_x}m$</span>. Put this expression for <span class="math-container">$x^\star$</span> into the expression for <span class="math-container">$y(x)$</span> from above to get <span class="math-container">$y^\star = \frac{1-\alpha}{p_y}m$</span>.</p>

CC BY-SA 4.0
2021-10-28T01:45:40.017Z
I've edited the question and fixed the typos, hopefully it's better now!
Lily Turner
CC BY-SA 4.0
2021-10-26T20:29:09.357Z
2021-10-26T20:29:09.357Z
chopschoc
<p>Derive the consumer's optimal demand of x and y, respectively, when the budget constraint is
<span class="math-container">$px + qy = m $</span>   and the utility function is`u = <span class="math-container">$x^p + y^p$</span></p>
<p>here's my attempt but i'm lost on how to solve for <span class="math-container">$x$</span>. Can someone please help with simple and easy to follow instructions?</p>
<p><a href="https://i.sstatic.net/EaulO.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/EaulO.jpg" alt="enter image description here" /></a></p>
<p>I'm lost after that. Please help me with steps. The answer for
<span class="math-container">$x=$</span> <span class="math-container">$((p/q))^((1/(p-1))$</span> * <span class="math-container">$y$</span>.</p>
<p>that's supposed to be power of <span class="math-container">$^(1/p-1)$</span> btw. i can solve the rest but idk where the power of <span class="math-container">$^(1/p-1)$</span> came from.</p>

CC BY-SA 4.0
2021-10-26T06:23:07.813Z
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.
Giskard
CC BY-SA 4.0
2021-10-27T19:58:38.720Z
@Giskard Hi, I fixed it. Hopefully it's better now.
Lily Turner
CC BY-SA 4.0
2021-10-28T10:00:16.870Z
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?
Giskard
<microeconomics>
Utility function and consumer's optimal demand (edited)
Lily Turner
# Utility function and consumer's optimal demand (edited)

(Asked by: username_0 on 2021-10-26)

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?

[![enter image description here](https://i.sstatic.net/EaulO.jpg)](https://i.sstatic.net/EaulO.jpg)

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.

**Tags:** microeconomics
> **Comments:**
> **username_1**: 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.  
> **username_0**: @Giskard Hi, I fixed it. Hopefully it's better now.  
> **username_1**: 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?  

---

## 1 answers

### Answer by username_2 on 2021-10-26. Score: 1 | Accepted answer

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
\begin{equation}
\max\_{x, y} \mathcal{L} = x^\alpha y^{1-\alpha}- \lambda\left(p\_xx + p\_yy - m\right).
\end{equation}

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 $y^\star = \frac{1-\alpha}{p\_y}m$.

> **Comments:**
> **username_0**: I've edited the question and fixed the typos, hopefully it's better now!  

---

