Today I see a problem very wired.
To find the maxima and minima of function
$$f(x,y) = x^3 – y^3 -12xy -63x + 63y$$
I tried to take the partial derivative with respect to x and y, and obtain:
$$\frac{\partial}{\partial x}f(x,y) = 3x^2 – 12y-63$$
$$\frac{\partial}{\partial y} = -3y^2 – 12x + 63$$
and when I tried to find the stationary point, I found that I don't know how to solve the system of equations.
Best Answer
My first advice is to learn latex my friend, it's very helpfull!
Lets consider $$f(x, y) = x^3 - y^3 - 12xy -63x + 63y$$ Taking both partial derivate of $x$ and $y$ and making them equal $0$, $$\frac{\partial}{\partial x} f(x, y) = 3x^2-12y-63 = 0, \quad (1)$$ and $$\frac{\partial}{\partial y} f(x, y) = -3y^2 -12x +63x = 0, \quad (2)$$ Clearing $y$ from equation $(1)$ and substituting in equation $(2)$, $$y=\frac{3x^2-63}{12} \Longrightarrow -3\Big(\frac{3x^2-63}{12}\Big)^2-12x+63=-\frac{3x^4}{16}+\frac{63x^2}{8}-12x+\Big(63-\frac{1323}{16}\Big)=0$$ which is a fourth degree polynomial, then $$x\in\{-7, -1, 3, 5\}$$ Using this set of points for $x$ on equation $(1)$ (or $(2)$, it's the same), were have the following set of stationary points, $$A = \{(-7, 7), (-1, -5), (3,-3), (5, 1)\}$$ Let's consider the following Theorem. Let $$D=\frac{\partial^2}{\partial x^2} f(a, b) \cdot \frac{\partial^2}{\partial y^2} f(a, b) -\Big(\frac{\partial^2}{\partial x \partial y} f(a, b)\Big)^2, \quad (a, b) \in A$$ then,
$1)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) > 0$, then $f$ has a minimum at $(a, b)$.
$2)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) < 0$, then $f$ has a maximum at $(a, b)$.
$3)$ If $D<0$, then $f$ has a saddle point at $(a, b)$.
$4)$ If $D=0$, then we have no conclusion for $f$ at $(a, b)$.
Using this Theorem with the points of the set $A$, we conclude that, $(-7, 7)$ is maximum of $f$, $(3, -3)$ is minimum of $f$ and both $(-1, -5)$ and $(5, 1)$ are saddle points of $f$.