F(xy) = (1/3)x^3 + 4xy - 9x - y^2?

Question

Find the local minimums, maximums and saddle points if any. Any help is appreciated. Thank you

Answer 1

The function F(x, y) = (1/3)x³ + 4xy - 9x - y² .

F_x(x, y) = x² + 4y - 9.

F_y(x, y) = 4x - 2y.

Solve the following equations  F_x = 0  and F_y = 0 simultaneously.

x² + 4y - 9 = 0

4x - 2y = 0 ⇒ 2x - y = 0 ⇒ y = 2x.

Substitute y = 2x in x² + 4y - 9 = 0.

x² + 4(2x) - 9 = 0

x² + 8x - 9 = 0

By factor by grouping.

x² + 9x - x - 9 = 0

x(x + 9) - 1(x + 9) = 0

(x + 9)(x - 1) = 0

x = - 9  and  x = 1.

When, x = - 9, y = 2(- 9) = - 18.

When, x = 1, y = 2(1) = 2.

The critical points are (1, 2) and (- 9, - 18).

Now find the second order partial derivatives.

F_xx(x, y) = 2x.

F_yy(x, y) = - 2.

F_xy(x, y) = 4.

If either x = 1 or y = 2, then d = F_xx(x, y)F_yy(x, y)   - [ F_xy(x, y) ]² .

= [2x * - 2] - 4²

= [2 * 1 * - 2] - 4²

= - 4 - 16

- 20 < 0.

At (1, 2), d < 0, this is a saddle point.

At (- 9, - 18), d = [2 * - 9 * - 2] - 4² = 36 - 16 = 20 > 0 and F_xx(x, y) = 2x = 2 * - 9 = - 18 < 0, hence, we have local maximum here.