Maximum and minimum of $f(x,y)=xy+x^2y^2$ for $\lvert x \rvert\le1$ and $\lvert y \rvert\le1$ Is there some 'clever' method to find the maximum and minimum of $f(x,y)=xy+x^2y^2$ for $\lvert x \rvert\le1$ and $\lvert y \rvert\le1$ using basic analysis (i.e. without partial derivative and stationery points)?
With $y=kx$, $k>0$ we have $0<x<x_0=\min(1,1/k)$ and $$g(x)=f(x,kx)=kx^2+k^2x^4=kx^2(1+kx^2)$$ which is an even function with $g'(x)=0$ for $x=0$ which gives $$\min(f)=g(0)=0$$ and $$\max(f)=g(x_0)=g(\min(1,1/k)).$$
For $0<k\le1$ we have $g(x_0)=g(1)=k(1+k)\le1\cdot2=2$.
For $k>1$ we have $g(x_0)=g(1/k)=\frac{1}{k}(1+\frac{1}{k^2})=\frac{1}{k}+\frac{1}{k^2}<2$.
So, so far $0\le f(x,y)\le 2$, and then one continues with $k<0$ etcetera. But, is there a quicker and 'more elegant' way of solving it?
(The answer is $\min(f)=-1/4$ and $\max(f)=2$.)
 A: Since $\lvert x \rvert\le1$ and $\lvert y \rvert\le1$ then $t:=xy\in [-1,1]$ and
$$f(x,y)=xy+x^2y^2=t(1+t)$$
is a quadratic function with respect to $t$. It should be easy to verify that in the interval $[-1,1]$ such quadratic function attains its maximum at $t=1$ and its minimum at $t=-1/2$. 
A: Yes: set $t=xy$. The function $f(x,y)$ is really the quadratic function of a single variable $g(t)=t+t^2$, and the hypotheses imply that $|t|\le 1$, so that you have to find the extrema of $g(t)$ on $[-1,1]$, which is a high school exercise.
A: Let  |$x|\le1$, and $|y| \le 1.$
Complete the square:
$f(x,y):=(xy+1/2)^2-1/4$.
1) Minimum:
$f(x,y)=$
$(xy+1/2)^2-1/4 \ge -1/4$.
$\min f(x,y) = -1/4$, for $xy =-1/2$;
2) Maximum:
Since $|x| \le 1$, and $|y| \le 1$, we have $xy \le 1$.
$\max f(x,y) =$
$(1+1/2)^2-1/4= 9/4-1/4=2$.
A: The maximum is actually easy, since $f(1,1)=2$ and, in your domain, $f(x,y) < 1+1=2$ for any $x<1$ or $y<1$
Also, if you can prove that the minimum is not in the "border", you can just take partial derivatives to find the minimum (getting very easily that $x=y$ and $y=-\frac{1}{2}$
