Finding absolute max and min values of function

Question

Function given as $f(x,y) = 3x^2 + 2xy^2$. If $(x,y)$ lies in the region inside including edges of the triangle in the first quadrant given by $x\ge0, y\ge0, y\le2-x$. Reduce $f$ to a single variable function on each side of the triangle.

Answer 1

1) Find the critical points in the interior of the triangle, by setting the partial derivatives equal to $0$.

2) Find the maxima, minima on the boundary, by using one variable techniques. Two of the line segments making up the boundary will be very easy to deal with.

3) Compare all the candidates, to determine the winner and the loser.

Answer 2

If this is homework and the method is specified, you may want to use that method. Else, given $f(x, y) = 3x^2 + 2xy^2$ and the feasible region, we have:

$y \le 2 - x \implies f(x, y) \le 3x^2 + 2x(2-x)^2$

So $f(x,y) \le 2x^3 - 5x^2 + 8x = 2x\left((x-\frac{5}{4})^2 + \frac{39}{16}\right)$

As this is increasing in $x$, we can maximise the function by choosing the largest feasible value of $x$, which is when $x = 2, y = 0$.

For the minimum, notice that $x \ge 0, y \ge 0 \implies$ each term in $3x^2 + 2xy^2$ is non-negative, hence the minimum is $f(x, y) = 0$, which would occur when $x=0$ for any allowed value for $y$.