<p>You need to be careful. You can apply the Lagrangian to your problem, but the solution you will obtain will not be a maximum but instead a minimum. The reason is that your objective function <span class="math-container">$x^2 + y^2$</span> is convex (and not concave).</p>
<p>If you are optimizing <span class="math-container">$x^2 + y^2$</span>, then the best thing to do is either to put all your money on purchasing <span class="math-container">$x$</span> or all your money on purchasing <span class="math-container">$y$</span>. The solution will be the following.</p>
<ol>
<li>if <span class="math-container">$p_x &lt; p_y$</span> then you will put all your money on buying <span class="math-container">$x$</span> (as you will be able to buy more of <span class="math-container">$x$</span> than of <span class="math-container">$y$</span>). As such, it will be optimal to set <span class="math-container">$x = \frac{m}{p_x}$</span> and to set <span class="math-container">$y$</span> equal to <span class="math-container">$0$</span>.</li>
<li>if <span class="math-container">$p_y &lt; p_x$</span> the reverse holds, so it will be optimal to set <span class="math-container">$x = 0$</span> and <span class="math-container">$y = \frac{m}{p_y}$</span>.</li>
<li>if <span class="math-container">$p_y = p_x$</span> then both previous cases apply. So you will either choose <span class="math-container">$x = \frac{m}{p_x}$</span> and <span class="math-container">$y = 0$</span> or you will choose <span class="math-container">$x = 0$</span> and <span class="math-container">$y = \frac{m}{p_y}$</span>.</li>
</ol>

CC BY-SA 4.0
2023-11-16T15:48:14.083Z
Thank you, of course, this makes sense. I should have checked the second order properties and I would noticed the solution I found was a minimum, and that the function was not concave.
Neil B
CC BY-SA 4.0
2023-11-17T00:13:06.543Z
Well, it is a minimum subject to being on the budget line. The minimum in the budget set is the bundle $(0,0)$.
Michael Greinecker
CC BY-SA 4.0
2023-11-15T17:34:43.353Z
2023-11-15T17:34:43.353Z
tdm
<p><a href="https://i.sstatic.net/YsdgY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YsdgY.png" alt="enter image description here" /></a></p>
<p>Observe that the points of tangency of the indifference curve and the budget line are not optimal, as these points lie on a lower indifference curve in all three cases compared to the indifference curve through the optimal choices.</p>
<p>Here is the demand:</p>
<p><span class="math-container">\begin{eqnarray*}(x^d,y^d)(p_X,p_Y, M)\in\begin{cases}\left\{\left(\dfrac{M}{p_X},0\right),\left(0,\dfrac{M}{p_Y}\right)\right\} &amp; \text{if } p_X=p_Y \\ \left\{\left(0,\dfrac{M}{p_Y}\right)\right\} &amp; \text{if } p_X&gt;p_Y \\ \left\{\left(\dfrac{M}{p_X},0\right)\right\} &amp; \text{if } p_X&lt;p_Y \end{cases} \end{eqnarray*}</span></p>

CC BY-SA 4.0
2023-11-16T11:43:41.810Z
A fourth graph illustrating show the suboptimal outcome arrived at using the tangency condition would be nice!
Giskard
CC BY-SA 4.0
2023-11-16T15:50:13.480Z
Thank you for the graphical explanation this helps with understanding the solution
Neil B
CC BY-SA 4.0
2023-11-16T04:15:58.823Z
2023-11-16T13:04:08.073Z
2023-11-16T13:04:08.073Z
Amit
<p>My question is regarding a simple marshallian demand calculation. Given a utility function <span class="math-container">$u(x,y)=x^2+y^2$</span> and a budget constraint <span class="math-container">$p_1x+p_2y=m$</span>. What are the Marshallian demand functions for each x and y?</p>
<p>I use the Lagrange method, divide the two efficiency equations by each other to get <span class="math-container">$x/y=p_1/p_2$</span> and <span class="math-container">$x=p_1y/x$</span>, substitute into the constraint equation to solve for <span class="math-container">$y=p_2m/p_1^2+p_2^2$</span> and then similarly <span class="math-container">$x=p_1m/p_1^2+p_2^2$</span></p>
<p>I think this is correct, however my tutor disagrees, they get as far as <span class="math-container">$x/y=p_1/p_2$</span>, then for some reason they say suppose <span class="math-container">$p_1=p_2=m=1$</span> then <span class="math-container">$x=y=1/2$</span> and in general <span class="math-container">$x=m/p_1 , y=m/p_2$</span>.</p>
<p>Can someone please let me know  which is correct and why you can just assume values for the prices and income when the question does not state this.</p>

CC BY-SA 4.0
2023-11-15T16:56:08.103Z
Both you and your tutor are on the wrong track. All optima will be corner solutions. Note that indifference curves are the northeast quadrants of circles with center at 0.
Michael Greinecker
<consumer-theory><demand>
Marshallian demand for x^2+y^2
Neil B
# Marshallian demand for x^2+y^2

(Asked by: username_0 on 2023-11-15)

My question is regarding a simple marshallian demand calculation. Given a utility function $u(x,y)=x^2+y^2$ and a budget constraint $p\_1x+p\_2y=m$. What are the Marshallian demand functions for each x and y?

I use the Lagrange method, divide the two efficiency equations by each other to get $x/y=p\_1/p\_2$ and $x=p\_1y/x$, substitute into the constraint equation to solve for $y=p\_2m/p\_1^2+p\_2^2$ and then similarly $x=p\_1m/p\_1^2+p\_2^2$

I think this is correct, however my tutor disagrees, they get as far as $x/y=p\_1/p\_2$, then for some reason they say suppose $p\_1=p\_2=m=1$ then $x=y=1/2$ and in general $x=m/p\_1 , y=m/p\_2$.

Can someone please let me know which is correct and why you can just assume values for the prices and income when the question does not state this.

**Tags:** consumer-theory, demand

> **Comments:**
> **username_1**: Both you and your tutor are on the wrong track. All optima will be corner solutions. Note that indifference curves are the northeast quadrants of circles with center at 0.  

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## 2 answers

### Answer by username_2 on 2023-11-15. Score: 1 | Accepted answer

You need to be careful. You can apply the Lagrangian to your problem, but the solution you will obtain will not be a maximum but instead a minimum. The reason is that your objective function $x^2 + y^2$ is convex (and not concave).

If you are optimizing $x^2 + y^2$, then the best thing to do is either to put all your money on purchasing $x$ or all your money on purchasing $y$. The solution will be the following.

1. if $p\_x < p\_y$ then you will put all your money on buying $x$ (as you will be able to buy more of $x$ than of $y$). As such, it will be optimal to set $x = \frac{m}{p\_x}$ and to set $y$ equal to $0$.
2. if $p\_y < p\_x$ the reverse holds, so it will be optimal to set $x = 0$ and $y = \frac{m}{p\_y}$.
3. if $p\_y = p\_x$ then both previous cases apply. So you will either choose $x = \frac{m}{p\_x}$ and $y = 0$ or you will choose $x = 0$ and $y = \frac{m}{p\_y}$.

> **Comments:**
> **username_0**: Thank you, of course, this makes sense. I should have checked the second order properties and I would noticed the solution I found was a minimum, and that the function was not concave.  
> **username_1**: Well, it is a minimum subject to being on the budget line. The minimum in the budget set is the bundle $(0,0)$.  

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### Answer by username_3 on 2023-11-16. Score: 2

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

Observe that the points of tangency of the indifference curve and the budget line are not optimal, as these points lie on a lower indifference curve in all three cases compared to the indifference curve through the optimal choices.

Here is the demand:

\begin{eqnarray\*}(x^d,y^d)(p\_X,p\_Y, M)\in\begin{cases}\left\{\left(\dfrac{M}{p\_X},0\right),\left(0,\dfrac{M}{p\_Y}\right)\right\} & \text{if } p\_X=p\_Y \\ \left\{\left(0,\dfrac{M}{p\_Y}\right)\right\} & \text{if } p\_X>p\_Y \\ \left\{\left(\dfrac{M}{p\_X},0\right)\right\} & \text{if } p\_X

> **Comments:**
> **username_4**: A fourth graph illustrating show the suboptimal outcome arrived at using the tangency condition would be nice!  
> **username_0**: Thank you for the graphical explanation this helps with understanding the solution  

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