Question

1. Check the function f(x,y)=x³-y³+9xy for it's extremum

Answer 1

To find an extremum, we should set both partial derivatives of the function equal to zero. That is

$\frac{\partial f(x,y)}{\partial x} = 3 x^2 + 9y = 0$ and $\frac{\partial f(x,y)}{\partial y} = -3 y^2 + 9x = 0$

From the first equation $y = -\frac{1}{3}x^2$ , substituting this into the second equation get $-\frac{x^4}{3}+9x = 0$ , or$x^4 - 27 x =0$. This equation has two solutions: x = 0 and x = 3, substituting them into expression for y we get two critical points (0, 0) and (3, -3)

Now calculate the second derivatives matrix

$\frac{\partial^2f(x,y)}{\partial x^2}= 6x$

$\frac{\partial^2f(x,y)}{\partial y^2}= -6y$

$\frac{\partial^2f(x,y)}{\partial x \partial y}= 9$

$D(x,y) = \frac{\partial^2f(x,y)}{\partial x^2} \frac{\partial^2f(x,y)}{\partial y^2} - (\frac{\partial^2f(x,y)}{\partial x \partial y})^2 = -36 x y - 81$

In point (0, 0), D(0, 0) = -81 < 0, so this point is a saddle point.

In point (3, -3) D(3, -3) = 324 - 81=243 >0, and $\frac{\partial^2f(x,y)}{\partial x^2} = 18 > 0$ , so in this point the function has minimum