How to solve following optimization problem?

[user_01]

The following is the mathematical expression for my model's rate expression. Variables $x,y$ are the controlling parameter, while the rest are positive constants.

$$\max_{x,y} \ ax + by^3 \ (s.t. \ 0\leq x \leq 1,\ 0\leq y\leq1)$$

Can I mathematically say that it is a convex problem within the limits of variables $x,y$?

The graph for the equation strictly follows the definition of convexity.
I have learned to solve the problems with KKT method, but I cannot understand how to resolve the inequality constraints $0 \leq (x,y) \leq 1$.

 

[I like Serena]

The following is the mathematical expression for my model's rate expression. Variables $x,y$ are the controlling parameter, while the rest are positive constants.
$$\max_{x,y} \ ax + by^3 \ (s.t. \ 0\leq x \leq 1,\ 0\leq y\leq1)$$
Can I mathematically say that it is a convex problem within the limits of variables $x,y$?
The graph for the equation strictly follows the definition of convexity.

Wiki explains that a convex problem is the optimization of a convex function over a convex set.
So it is indeed a convex problem as both conditions are satisfied.

I have learned to solve the problems with KKT method, but I cannot understand how to resolve the inequality constraints $0 \leq (x,y) \leq 1$.

Those inequality constraints translate into the constraint function $\mathbf g(x,y) = (-x,-y,x-1,y-1)$ with $g_i(x,y) \le 0$.

Btw, we can already see by inspection that the optimum must be at $(x,y)=(1,1)$ with value $a+b$.

 

[HOI]

The derivative of $ax+ by^3$ with respect to x is the constant a. If a is not 0, that derivative is never 0 so any max or min must be on the boundary. That boundary consists of 4 pieces

1) x= 0, $0\le y\le 1$. The function reduces to $by^3$ on $0\le y\le 1$ which obviously has its minimum value, 0, at y= 0 and maximum value, b, at y= 1.
2) y= 0, $0\le x\le 1$. The function reduces to $ax$ on $0\le x\le 1$ which obviouslly has its minimum value, 0, at x= 0 and maximum value , a, a x=1.
3) x= 1, $0\le y\le 1$. The function reduces to $a+ by^3$ on $0\le y\le 1$ which obviouly has its minimum, a, at y= 0 and its maximum a+ b at y= 1.
4. y= 1, $0\le x\le 1$. The function reduces to $ax+ b$ on $0\le x\le 1$ which obviouslly has its minimum, b, at x= 0 and its maximum, a+ b at x= 1.

Over all, then, the maximum is a+ b which is achieved at (1, 1). The minimum value is 0 which is achieved at (0, 0).