What is the maximum of the following function?

Question

Let $f(x,y) = \frac{xy^\alpha}{x+y},\alpha\in(0,\infty)$. How to compute $$\sup_{(x,y)\in[a,b]\times [0,c]}\frac{xy^\alpha}{x+y},$$ with $b>a>0$?

Answer 1

Hint

Note that $$\dfrac{\partial f}{\partial x}=\dfrac{y^{\alpha+1}}{(x+y)^2}>0.$$ Thus, if $x>x_0$ then $f(x,y_0)>f(x_0,y_0).$ So, the maximum point is of the form $(b,y).$ But, on those points it is

$$g(y)=f(b,y)=\dfrac{by^{\alpha}}{b+y}.$$ You only need to maximize the function $g(y)$ in $ [0,c].$