<p>Intuitively, one can see that the corner solutions lie on the highest possible indifference curve(for concave preferences). Thus, utility obtained will be highest on those points. The following figure(courtesy-<a href="https://slideplayer.com/slide/9320004/" rel="nofollow noreferrer" title="Google Images">Google Images</a>) might help one to visualize the above statement.</p>
<p><a href="https://i.sstatic.net/dINWb.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dINWb.jpg" alt="enter image description here" /></a></p>
<p>Mathematically, it is a bit tough to visualize whats going on. However, the Kuhn Tucker conditions can tell you whether a corner solution will exist or not(note that concavity/convexity of preferences are of no use while using Kuhn-Tucker Theorem, however one needs to check for differentiability of the objective function/constraints). First of all, according to Weierstrass' Theorem, it is guaranteed that an optimum point for <span class="math-container">$U(x_1,x_2)$</span> exists in the budget set <span class="math-container">$B(p_1,p_2,w)=\{(x_1,x_2):p_1 x_1 + p_2 x_2 \leq w\}$</span> (only if <span class="math-container">$U(x_1,x_2)$</span> is continuous in the given domain,i.e; the budget set). This tells us that we cannot rule out any solution/optimal point where either <span class="math-container">$x_1=0$</span> or <span class="math-container">$x_2=0$</span>, or both. Now, to check whether a corner solution exists or not, we can proceed as follows. Consider the maximization problem</p>
<p><span class="math-container">\begin{gather*}
\max\ U(x_1,x_2); \\ s.t.\ p_1x_1+p_2x_2\leq w;\ x_1\geq 0, x_2\geq 0.\\
\end{gather*}</span>
The Kuhn Tucker Lagrangian is given as
<span class="math-container">\begin{align}
L = U(x_1,x_2)+ \lambda(w-p_1x_1+p_2x_2)+ \mu_1 x_1+ \mu_2 x_2.
\end{align}</span></p>
<p>The Kuhn Tucker conditions are given as-
<span class="math-container">\begin{gather*}
\frac{\partial L}{\partial x_1}\leq0\tag{1}\\ \frac{\partial L}{\partial x_2}\leq0\tag{2}\\ \frac{\partial L}{\partial \lambda}\leq0 ,\ \frac{\partial L}{\partial \mu_1}\leq0\ ,\frac{\partial L}{\partial \mu_2}\leq0\tag{3}\\ \lambda\frac{\partial L}{\partial \lambda}=0\tag{4}\\ \mu_1\frac{\partial L}{\partial \mu_1}=0\ ,\mu_2\frac{\partial L}{\partial \mu_2}=0\tag{5}
\end{gather*}</span></p>
<p>The inequality signs in conditions <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> can be replaced with the equality sign, if the constraints are binding(you can read more about binding constraints/constraint qualifications from Simon and Blume).</p>
<p>However, the given problem of corner solutions can be understood from the conditions in <span class="math-container">$(5)$</span>, which say that <span class="math-container">$\mu_1 x_1=0$</span> and <span class="math-container">$\mu_2 x_2=0$</span>. These equations tell us that either any one(or both) of the <span class="math-container">$\mu_i=0$</span> or any one(or both) of the <span class="math-container">$x_i = 0$</span>. By considering all the above cases one by one, we can figure out whether a corner solution will exist or not.</p>
<p>For example, if the utility function is given as <span class="math-container">$U(x_1,x_2) = x_1 + \ln(x_2)$</span>, the solutions obtained by using Kuhn Tucker theorem are-</p>
<p><span class="math-container">\begin{gather*}
    x_1= 
\begin{cases}
    \frac{w}{p_1} - 1,&amp; \text{if } w&gt;p_1\\
    0,              &amp; \text{otherwise}
\end{cases}\\
x_2= 
\begin{cases}
    \frac{p_1}{p_2} ,&amp; \text{if } w&gt;p_1\\
    \frac{w}{p_2},              &amp; \text{otherwise}
\end{cases}
\end{gather*}</span></p>
<p>The corner solutions in this case are obtained whenever <span class="math-container">$w\leq p_1$</span>.</p>

CC BY-SA 4.0
2018-09-28T15:28:33.503Z
2018-09-28T19:00:18.893Z
2020-06-17T09:40:54.643Z
superhulk
<p>I seem to not be getting this. Could someone explain me the mathematical way to show a concave utility function [like (ax^2+by^2)] subject to a budget constraint has a corner solution. I get the economic intuition behind it but not the mathematical part behind it. </p>

CC BY-SA 4.0
2018-09-27T15:50:25.643Z
$ax^2+by^2$ is *not* concave, unless $a,b<0$.
Herr K.
<optimization>
Concave utility functions corner solution explanation
Sumukh Sai
# Concave utility functions corner solution explanation

(Asked by: username_0 on 2018-09-27)

I seem to not be getting this. Could someone explain me the mathematical way to show a concave utility function [like (ax^2+by^2)] subject to a budget constraint has a corner solution. I get the economic intuition behind it but not the mathematical part behind it.

**Tags:** optimization
> **Comments:**
> **username_1**: $ax^2+by^2$ is *not* concave, unless $a,b<0$.  

---

## 1 answers

### Answer by username_2 on 2018-09-28. Score: 2 | Accepted answer

Intuitively, one can see that the corner solutions lie on the highest possible indifference curve(for concave preferences). Thus, utility obtained will be highest on those points. The following figure(courtesy-[Google Images](https://slideplayer.com/slide/9320004/ "Google Images")) might help one to visualize the above statement.

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

Mathematically, it is a bit tough to visualize whats going on. However, the Kuhn Tucker conditions can tell you whether a corner solution will exist or not(note that concavity/convexity of preferences are of no use while using Kuhn-Tucker Theorem, however one needs to check for differentiability of the objective function/constraints). First of all, according to Weierstrass' Theorem, it is guaranteed that an optimum point for $U(x\_1,x\_2)$ exists in the budget set $B(p\_1,p\_2,w)=\{(x\_1,x\_2):p\_1 x\_1 + p\_2 x\_2 \leq w\}$ (only if $U(x\_1,x\_2)$ is continuous in the given domain,i.e; the budget set). This tells us that we cannot rule out any solution/optimal point where either $x\_1=0$ or $x\_2=0$, or both. Now, to check whether a corner solution exists or not, we can proceed as follows. Consider the maximization problem

\begin{gather\*}
\max\ U(x\_1,x\_2); \\ s.t.\ p\_1x\_1+p\_2x\_2\leq w;\ x\_1\geq 0, x\_2\geq 0.\\
\end{gather\*}
The Kuhn Tucker Lagrangian is given as
\begin{align}
L = U(x\_1,x\_2)+ \lambda(w-p\_1x\_1+p\_2x\_2)+ \mu\_1 x\_1+ \mu\_2 x\_2.
\end{align}

The Kuhn Tucker conditions are given as-
\begin{gather\*}
\frac{\partial L}{\partial x\_1}\leq0\tag{1}\\ \frac{\partial L}{\partial x\_2}\leq0\tag{2}\\ \frac{\partial L}{\partial \lambda}\leq0 ,\ \frac{\partial L}{\partial \mu\_1}\leq0\ ,\frac{\partial L}{\partial \mu\_2}\leq0\tag{3}\\ \lambda\frac{\partial L}{\partial \lambda}=0\tag{4}\\ \mu\_1\frac{\partial L}{\partial \mu\_1}=0\ ,\mu\_2\frac{\partial L}{\partial \mu\_2}=0\tag{5}
\end{gather\*}

The inequality signs in conditions $(1)$ and $(2)$ can be replaced with the equality sign, if the constraints are binding(you can read more about binding constraints/constraint qualifications from Simon and Blume).

However, the given problem of corner solutions can be understood from the conditions in $(5)$, which say that $\mu\_1 x\_1=0$ and $\mu\_2 x\_2=0$. These equations tell us that either any one(or both) of the $\mu\_i=0$ or any one(or both) of the $x\_i = 0$. By considering all the above cases one by one, we can figure out whether a corner solution will exist or not.

For example, if the utility function is given as $U(x\_1,x\_2) = x\_1 + \ln(x\_2)$, the solutions obtained by using Kuhn Tucker theorem are-

\begin{gather\*}
x\_1=
\begin{cases}
\frac{w}{p\_1} - 1,& \text{if } w>p\_1\\
0, & \text{otherwise}
\end{cases}\\
x\_2=
\begin{cases}
\frac{p\_1}{p\_2} ,& \text{if } w>p\_1\\
\frac{w}{p\_2}, & \text{otherwise}
\end{cases}
\end{gather\*}

The corner solutions in this case are obtained whenever $w\leq p\_1$.


---

