<p>Your Lagrangian would be
<span class="math-container">$$L = (ax+by)+\lambda (I−p_x x−p_y y) +\mu_x(x−0)+\mu_y(y-0),$$</span>
where the final two terms represent the restriction that <span class="math-container">$x,y\geq0$</span>.</p>
<p>You then arrive at conditions
<span class="math-container">$$\frac{\partial L}{\partial x}= a -\lambda p_x +\mu_x=0$$</span>
<span class="math-container">$$\frac{\partial L}{\partial y}= b -\lambda p_y +\mu_y=0$$</span>
<span class="math-container">$$I=p_x x+p_y y$$</span>
and complementary slackness conditions
<span class="math-container">$$\mu_x x =0 , \quad \mu_y y =0.$$</span></p>
<p>You then consider many different cases: <span class="math-container">$(\mu_x&gt;0,x=0,\mu_y=0,y&gt;0),(\mu_x=0,x&gt;0,\mu_y=0,y&gt;0), (\mu_x=0,x=0,\mu_y=0,y&gt;0), ....$</span>
and for each of them you look at your conditions:</p>
<p>Suppose <span class="math-container">$(\mu_x&gt;0,x=0,\mu_y=0,y&gt;0)$</span>. Then,
<span class="math-container">$$a -\lambda p_x +\mu_x=0 \Rightarrow a -\lambda p_x &lt;0$$</span>
<span class="math-container">$$b -\lambda p_y +\mu_y=b -\lambda p_y=0$$</span>
<span class="math-container">$$\Rightarrow a -\frac{b}{p_y} p_x &lt;0 \iff \frac{a}{p_x} &lt;\frac{b}{p_y}$$</span>
This condition can be checked as <span class="math-container">$a,b,p_x,p_y$</span> are given. If it does not hold, <span class="math-container">$x=0,y&gt;0$</span> cannot be the solution. If it does hold, it means that the utility per dollar is higher for good <span class="math-container">$y$</span> and all money should be spend on <span class="math-container">$y$</span> (which <span class="math-container">$x=0,y&gt;0$</span> and the budget constraint imply).</p>
<p>You have to check all the cases, but you will arrive at a solution of the above form.</p>

CC BY-SA 4.0
2020-12-11T22:31:01.660Z
2020-12-11T22:31:01.660Z
Bayesian
<p>How does one solve utility maximization of perfect substitutes using Lagrangian function?</p>
<p>Consider the problem
<span class="math-container">$$\max_{x,y} ax +by  $$</span></p>
<p>subject to the constraint that</p>
<p><span class="math-container">$$px + qy \leq I$$</span></p>
<p>where <span class="math-container">$a,b,p,q,I&gt;0$</span>.</p>
<p>Note problem statement are subject to the standard implicit assumptions that consumption cannot be negative.</p>

<microeconomics><utility><optimization><demand><perfect-substitutes>
Perfect substitutes and Lagrange
Jesper Hybel
# Perfect substitutes and Lagrange

(Asked by: username_0 on 2020-12-11)

How does one solve utility maximization of perfect substitutes using Lagrangian function?

Consider the problem
$$\max\_{x,y} ax +by $$

subject to the constraint that

$$px + qy \leq I$$

where $a,b,p,q,I>0$.

Note problem statement are subject to the standard implicit assumptions that consumption cannot be negative.

**Tags:** microeconomics, utility, optimization, demand, perfect-substitutes

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## 1 answers

### Answer by username_1 on 2020-12-11. Score: 6 | Accepted answer

Your Lagrangian would be
$$L = (ax+by)+\lambda (I−p\_x x−p\_y y) +\mu\_x(x−0)+\mu\_y(y-0),$$
where the final two terms represent the restriction that $x,y\geq0$.

You then arrive at conditions
$$\frac{\partial L}{\partial x}= a -\lambda p\_x +\mu\_x=0$$
$$\frac{\partial L}{\partial y}= b -\lambda p\_y +\mu\_y=0$$
$$I=p\_x x+p\_y y$$
and complementary slackness conditions
$$\mu\_x x =0 , \quad \mu\_y y =0.$$

You then consider many different cases: $(\mu\_x>0,x=0,\mu\_y=0,y>0),(\mu\_x=0,x>0,\mu\_y=0,y>0), (\mu\_x=0,x=0,\mu\_y=0,y>0), ....$
and for each of them you look at your conditions:

Suppose $(\mu\_x>0,x=0,\mu\_y=0,y>0)$. Then,
$$a -\lambda p\_x +\mu\_x=0 \Rightarrow a -\lambda p\_x <0$$
$$b -\lambda p\_y +\mu\_y=b -\lambda p\_y=0$$
$$\Rightarrow a -\frac{b}{p\_y} p\_x <0 \iff \frac{a}{p\_x} <\frac{b}{p\_y}$$
This condition can be checked as $a,b,p\_x,p\_y$ are given. If it does not hold, $x=0,y>0$ cannot be the solution. If it does hold, it means that the utility per dollar is higher for good $y$ and all money should be spend on $y$ (which $x=0,y>0$ and the budget constraint imply).

You have to check all the cases, but you will arrive at a solution of the above form.


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