<p>Because the two segments of the IC bend away from the origin, the only possible solutions are at the corners or at <span class="math-container">$x=y$</span>.</p>

<p>Write <span class="math-container">$p_x$</span> and <span class="math-container">$p_y$</span> for the prices, and <span class="math-container">$M$</span> for the budget.</p>

<p>Suppose we try to set up a corner solution in which only one of the two goods is consumed. Since utility is symmetric, it makes sense to spend all of your money on whichever good is cheaper. Suppose this is good <span class="math-container">$x$</span>. We then have <span class="math-container">$x=\frac{M}{p_x}&gt;0=y$</span> and <span class="math-container">$u=x$</span>.</p>

<p>The best <span class="math-container">$x=y$</span> bundle that can be afforded is such that <span class="math-container">$x p_x+x p_y=M\iff x=y=\frac{M}{p_x+p_y}$</span>. Utility is then
<span class="math-container">$$\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y}.$$</span></p>

<p>You should easily be able to compute a condition on <span class="math-container">$M$</span> such that the bundle with <span class="math-container">$x=y$</span> is optimal: <span class="math-container">$$M&gt;\frac{p_y(p_x+p_y)}{p_x}.$$</span></p>

<p>You can see this dependence on M if we look at a contour plot of the indifference curves:</p>

<p><a href="https://i.sstatic.net/ytC8E.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ytC8E.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2018-11-27T10:30:23.167Z
Thanks for the nice explanation with the graph included. Btw, so is the case of $x=y$ an interior solution? From the graph it appears as though you can have interior solution at $x=y$?
OGC
CC BY-SA 4.0
2018-11-27T10:33:35.893Z
@OGC It depends what you mean by 'interior solution' (the definition of that term is a bit fuzzy). The solution is interior in the sense that both goods are consumed in positive quantities (so long as $M$ is large enough), but non-interior in the sense that the Marginal Rate of Substitution does not equal the relative price as in the typical textbook solution to utility maximization.
Ubiquitous
CC BY-SA 4.0
2018-11-27T11:28:20.600Z
@Ubiquitious I see. By interior I meant $x$ and $y$ being both positive. However, I do not quite follow how you were able to obtain the price condition for $x=y$ case. Normally, I  would try to find such a condition from the indirect utility function, $v\left ( p,u \right ) = max \left \{ \frac{M}{p_{x}}, \frac{M}{p_{y}},\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y} \right \}$.
OGC
CC BY-SA 4.0
2018-11-27T12:35:12.873Z
@OGC The budget constraint is $xp_x+yp_y=M$. I am looking for the best bundle with $x=y$, so I make this substitution to yield $xp_x+xp_y=M$. Then I simply solve to find the quantity $x$ that expends the full income (the quantity of $y$ in the bundle is the same by definition).
Ubiquitous
CC BY-SA 4.0
2018-11-27T12:38:01.600Z
@OGC Or did you mean the last condition? That is found by setting $\max_x u(x,0)<\max_x u(x,x)$ and solving for $M$: $$\frac{M}{p_X}<\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y}\implies M>\frac{p_y(p_x+p_y)}{p_x}$$.
Ubiquitous
CC BY-SA 4.0
2018-11-27T09:41:09.587Z
2018-11-27T09:41:09.587Z
Ubiquitous
<p>The above answer needs to be supplemented with this answer. You can't just say "the graph looks concave". You have to put in a little work, at least if you were my student.</p>

<p>Let <span class="math-container">$x&gt; y$</span> without loss of generality (the objective function is symmetrical). Then we have <span class="math-container">$U(x,y) = y^2+x$</span>. Since this is increasing in both variables, we can see right away that our constraint (<span class="math-container">$x+py \leq M$</span>) will bind (you can provide an epsilon delta argument for this if you want). Thus we have <span class="math-container">$U(x(y),y) = y^2+M-py$</span>. This is a polynomial, and we find that <span class="math-container">$y = 2/p$</span> is the only critical point and by the second derivative test it is a local minimum.</p>

<p>Now we have to consider the case <span class="math-container">$x=y$</span>. This is what the answer above me can be inserted.</p>

<p>Have a good day. Oh and <span class="math-container">$x=y$</span> is most definitely an interior point in this context. Although technically that solution is on the boundary of the choice set. The only reason it is a special case is that the function isn't differentiable there.</p>

CC BY-SA 4.0
2018-11-27T17:48:24.033Z
2018-11-27T18:19:54.143Z
2018-11-27T18:19:54.143Z
adca
<p>Will all the solutions be in the corner or will the cusp in the middle give us any interior solution? This is by the intersection of the budget line.</p>

<p>I am getting this type of a shape:
<a href="https://i.sstatic.net/GTPY5.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GTPY5.png" alt="enter image description here"></a></p>

<p>But I am not sure if there will be interior solution. If there is we can't use differentiation because of the cusp. I think that there will only be corner solutions but it depends on the shape of the indifference curve. </p>

<utility><optimization>
Any interior solution for $u(x,y) = min\left \{ x,y \right \}^{2} + max\left \{ x,y \right \}$?
OGC
# Any interior solution for $u(x,y) = min\left \{ x,y \right \}^{2} + max\left \{ x,y \right \}$?

(Asked by: username_0 on 2018-11-27)

Will all the solutions be in the corner or will the cusp in the middle give us any interior solution? This is by the intersection of the budget line.

I am getting this type of a shape:
[![enter image description here](https://i.sstatic.net/GTPY5.png)](https://i.sstatic.net/GTPY5.png)

But I am not sure if there will be interior solution. If there is we can't use differentiation because of the cusp. I think that there will only be corner solutions but it depends on the shape of the indifference curve.

**Tags:** utility, optimization

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## 2 answers

### Answer by username_1 on 2018-11-27. Score: 1 | Accepted answer

Because the two segments of the IC bend away from the origin, the only possible solutions are at the corners or at $x=y$.

Write $p\_x$ and $p\_y$ for the prices, and $M$ for the budget.

Suppose we try to set up a corner solution in which only one of the two goods is consumed. Since utility is symmetric, it makes sense to spend all of your money on whichever good is cheaper. Suppose this is good $x$. We then have $x=\frac{M}{p\_x}>0=y$ and $u=x$.

The best $x=y$ bundle that can be afforded is such that $x p\_x+x p\_y=M\iff x=y=\frac{M}{p\_x+p\_y}$. Utility is then
$$\frac{M^2}{(p\_x+p\_y)^2}+\frac{M}{p\_x+p\_y}.$$

You should easily be able to compute a condition on $M$ such that the bundle with $x=y$ is optimal: $$M>\frac{p\_y(p\_x+p\_y)}{p\_x}.$$

You can see this dependence on M if we look at a contour plot of the indifference curves:

[![enter image description here](https://i.sstatic.net/ytC8E.png)](https://i.sstatic.net/ytC8E.png)

> **Comments:**
> **username_0**: Thanks for the nice explanation with the graph included. Btw, so is the case of $x=y$ an interior solution? From the graph it appears as though you can have interior solution at $x=y$?  
> **username_1**: @OGC It depends what you mean by 'interior solution' (the definition of that term is a bit fuzzy). The solution is interior in the sense that both goods are consumed in positive quantities (so long as $M$ is large enough), but non-interior in the sense that the Marginal Rate of Substitution does not equal the relative price as in the typical textbook solution to utility maximization.  
> **username_0**: @Ubiquitious I see. By interior I meant $x$ and $y$ being both positive. However, I do not quite follow how you were able to obtain the price condition for $x=y$ case. Normally, I  would try to find such a condition from the indirect utility function, $v\left ( p,u \right ) = max \left \{ \frac{M}{p_{x}}, \frac{M}{p_{y}},\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y} \right \}$.  
> **username_1**: @OGC The budget constraint is $xp_x+yp_y=M$. I am looking for the best bundle with $x=y$, so I make this substitution to yield $xp_x+xp_y=M$. Then I simply solve to find the quantity $x$ that expends the full income (the quantity of $y$ in the bundle is the same by definition).  
> **username_1**: @OGC Or did you mean the last condition? That is found by setting $\max_x u(x,0)<\max_x u(x,x)$ and solving for $M$: $$\frac{M}{p_X}<\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y}\implies M>\frac{p_y(p_x+p_y)}{p_x}$$.  

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### Answer by username_2 on 2018-11-27. Score: 1

The above answer needs to be supplemented with this answer. You can't just say "the graph looks concave". You have to put in a little work, at least if you were my student.

Let $x> y$ without loss of generality (the objective function is symmetrical). Then we have $U(x,y) = y^2+x$. Since this is increasing in both variables, we can see right away that our constraint ($x+py \leq M$) will bind (you can provide an epsilon delta argument for this if you want). Thus we have $U(x(y),y) = y^2+M-py$. This is a polynomial, and we find that $y = 2/p$ is the only critical point and by the second derivative test it is a local minimum.

Now we have to consider the case $x=y$. This is what the answer above me can be inserted.

Have a good day. Oh and $x=y$ is most definitely an interior point in this context. Although technically that solution is on the boundary of the choice set. The only reason it is a special case is that the function isn't differentiable there.


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