Find the maximum value of E(x,y) [closed]

First of all, as KingW3 said, we can assume $x \ge y$ without loss of generality since $E(x, y) = E(y, x)$. Which gives

$$E(x, y) = xy + x = x(y+1)$$

Then, I agree on the trigonometric approach, and because $x^2 + y^2 = 1$, I suggest to replace $x$ with $\cos(\theta)$ and $y$ with $\sin(\theta)$, which permits to reduce E(x, y) to only one variable, the angle. Let's call $f(\theta) = E(\cos(\theta), \sin(\theta))$.

$$f(\theta) = \cos(\theta)(\sin(\theta) + 1)$$

Our goal is thus to maximize $f(\theta)$, which we can do with calculus. We first search for $f'(\theta)$

$$f'(\theta) = -\sin(\theta)(\sin(\theta) + 1) + \cos(\theta)\cos(\theta)$$

So $f'(\theta) = \cos(\theta)^2 -\sin(\theta)^2 - \sin(\theta)$ and we know that $\cos(\theta)^2 = 1 - \sin(\theta)^2$, allowing us to rewrite

$$f'(\theta) = 1 - \sin(\theta)^2 - \sin(\theta)$$

From there, we want to search for $f'(\theta) = 0 \Leftrightarrow 1 - \sin(\theta)^2 - \sin(\theta) = 0$. Then, even if it not obvious right away, we can try to factorize what we just obtained, which can be done in the following way:

$$f'(\theta) = -(\sin(\theta) + 1)(2\sin(\theta) - 1)$$

So we want at least one of the two parts to be 0 for the overall to be 0. But, since we want $x \ge 0$ and $y \ge 0$, on the range from $0$ to $2\pi$ we must have $0 \ge \theta \ge \frac{\pi}{2}$, and that means that $\sin(\theta) + 1$ has no solution over that range. Our problem thus reduces only to

$$2sin(\theta) - 1 = 0 \Leftrightarrow \sin(\theta) = y = \frac{1}{2}$$

From there, we deduce

$$x^2 + y^2 = 1 \Leftrightarrow x = \sqrt{1 - y^2} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$$

And with this, we obtain our final solution,

$$\max(E(x, y)) = E\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) = \frac{\sqrt{3}}{2}\times \left(\frac{1}{2} + 1\right) = \frac{3\sqrt{3}}{4}$$