From reading online I found someone said that it has a unique optimal solution when $a$ and $b$ are positive and $a \neq b$.
Could someone explain why this is the case?
I know that if $a = b$ then any x,y values that satisfy the equation $x + y = (1/a)$ are optimal, and thus the solution is not unique. But I am confused about how to show that the solution for when $a < b$ and when $a > b$ are unique.
It appears that you want to maximize $f(x,y) = x+y$ subject to constraints $x\ge 0$, $y\ge0$, $ax+by\le 1$.
If $0<a<b$ then it is easily seen that $f(\frac1a,0)=\frac1a$ is the unique optimal solution. Indeed, $$x+y\le \frac1a(ax+ay)+\frac{b-a}ay=\frac1a(ax+by)=\frac1a$$ with equality only if $\frac{b-a}ay=0$, i.e. $y=0$.
If $0<b<a$, similarly $f(0,\frac1b)=\frac1b$ is the unique optimom.
If $0<a=b$ then $f(x,y)=x+y\le \frac1a(ax+ay)=\frac1a$ with equality for several points, namely $f(t,\frac1a-t)=\frac1a$ for $0\le t \le \frac1a$.
If $a\le 0$, then $x$ is allowed to be arbitrarily large, hence $f(x,0)=x$ becomes arbitrarily large and there is no optimum.
Similarly if $b\le0$ with $f(0,y)$.
Draw a picture of the set $ax + by \le 1$, $x \ge 0$, $y \ge 0$. (For $a$ and $b$ positive, this is a triangle in the first quadrant.)
The function $f(x,y) = x+y$ is constant on lines $x+y = c$. Change $c$ and you will get a family of parallel lines. Adjust $c$ so that it is as large as possible and still touches the triangle.
