Local extrema $f(x, y) = x^3 + 3xy^2 − 39x − 36y + 26$.

Question

I am looking for the local extrema of

$$f(x, y) = x^3 + 3xy^2 − 39x − 36y + 26$$

but I don't know how to proceed.

I was trying even finding the derivative with respect to $x$ then with respect to $y$ but didn't get any result.

Any help?

Answer 1

In the extrema the partials wrt $x$ and wrt $y$ have to be both zero

$\dfrac{\partial f}{\partial x}=3x^2+3y^2-39=0$

$\dfrac{\partial f}{\partial y}=6xy-36=0$

And solve the resulting system of equations. This will give the stationary points for later find out which of them are extrema.

For $x$, $x^4-13x^2+36=0$. With $t=x^2$, we get $t=4$ or $t=9$, $x=\pm2$ or $x=\pm3$

Using the previous relation of $xy=6$, the points are:

$(2,3)\;;(-2-3)\;;(3,2)\;;(-3,-2)$

To know the type of each stationary point we have to evaluate the hessian, with the second partials:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-f_{xy}(x,y)^2$

$D(x,y)=36x^2-36y^2$

$D(2,3)\lt0$ then it's a saddle point.

$D(-2,-3)\lt0$ then it's a saddle point.

$D(3,2)\gt0$ and $f_{xx}\gt0$, minimum.

$D(-3,-2)\gt0$ and $f_{xx}\lt0$, maximum