Global extreme points of function wrt. two variables

Solution 1:

We will use these extrema in two varaibles notes as a guide.

We have

$$U(x, y) = (x\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2+3\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right)$$

Finding critical points

$$f_x = \left(x^3-x^2-2 x\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right) = 0\\ f_y = \left(\frac{x^4}{4}-\frac{x^3}{3}-x^2+3\right) \left(y^3-y^2-2 y\right) = 0$$

From the first, we have $x = -1, 0, 2$ and from the second, we have $y = -1,0,2$ to give us a total of nine critical points as

$$(x, y) = (-1,-1),(-1,0),(-1,2),(0,-1),(0,0),(0,-2),(2,-1),(2,0),(2,2)$$

The Hessian determinant, $|H(x, y)|$ is given by

$\left|\left( \begin{array}{cc} \left(3 x^2-2 x-2\right) \left(\dfrac{y^4}{4}-\dfrac{y^3}{3}-y^2+4\right) & \left(x^3-x^2-2 x\right) \left(y^3-y^2-2 y\right) \\ \left(x^3-x^2-2 x\right) \left(y^3-y^2-2 y\right) & \left(\dfrac{x^4}{4}-\dfrac{x^3}{3}-x^2+3\right) \left(3 y^2-2 y-2\right) \\ \end{array} \right)\right|$

We also have

$$f_{xx} (x, y) = \left(3 x^2-2 x-2\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right)$$

For the nine critical points, we find

• Saddles: $(-1,0), (0, -1), (0, 2), (2, 0)$

• Local Min: $(-1, 1), (-1, 2), (2, -1)$

• Local Max: $(0, 0)$

• Global Min: $(2,2)$