Lagrange Multiplier

In this simple example, it may be easiest to solve the constraint for $y$ . We have $x^2y^2=\frac{1}{4}x^2(2x-x^2)$ . Setting the derivative to zero, $\frac{3}{2}x^2-x^3=0$ , shows that the maximum is attained at $x=\frac{3}{2}$ . The maximal value of $x^2y^2$ is $\frac{27}{64}$ , and the maximum of $xy$ is $\frac{3\sqrt{3}}{8}\approx \boxed{0.650}$ .

It's equivalent to the ellipse:

(x-1)² + (y-1)²/(1/4) = 1

Then let

x -1 = cosΦ x = cosΦ +1

y/(1/2) = sinΦ y = (1/2)sinΦ

Then do :

xy = (1+cosΦ)[(1/2)sinΦ]

Take then the derivative and equate to zero.