# Marshallian demand for x^2+y^2

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.

• 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. Commented Nov 15, 2023 at 16:56

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 $$y = \frac{m}{p_y}$$.
• 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. Commented Nov 16, 2023 at 15:48
• Well, it is a minimum subject to being on the budget line. The minimum in the budget set is the bundle $(0,0)$. Commented Nov 17, 2023 at 0:13

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

• A fourth graph illustrating show the suboptimal outcome arrived at using the tangency condition would be nice! Commented Nov 16, 2023 at 11:43
• Thank you for the graphical explanation this helps with understanding the solution Commented Nov 16, 2023 at 15:50