Perfect substitutes and Lagrange 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 $a,b,p,q,I&gt;0$.
Note problem statement are subject to the standard implicit assumptions that consumption cannot be negative.
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: $(\mu_x&gt;0,x=0,\mu_y=0,y&gt;0),(\mu_x=0,x&gt;0,\mu_y=0,y&gt;0), (\mu_x=0,x=0,\mu_y=0,y&gt;0), ....$
and for each of them you look at your conditions:
Suppose $(\mu_x&gt;0,x=0,\mu_y=0,y&gt;0)$. Then,
$$a -\lambda p_x +\mu_x=0 \Rightarrow a -\lambda p_x &lt;0$$
$$b -\lambda p_y +\mu_y=b -\lambda p_y=0$$
$$\Rightarrow a -\frac{b}{p_y} p_x &lt;0 \iff \frac{a}{p_x} &lt;\frac{b}{p_y}$$
This condition can be checked as $a,b,p_x,p_y$ are given. If it does not hold, $x=0,y&gt;0$ 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 $x=0,y&gt;0$ and the budget constraint imply).
You have to check all the cases, but you will arrive at a solution of the above form.