Question

Obtain and classify the relative extrema values of the function
F(x,y)=x³+y³-63(x+y)+12xy

Answer 1

$F(x,y)=x^3+y^3-63(x+y)+12xy$

0 It's a polynomial function with continuous partial derivatives of any order.

1 What are the critical points?

$F^{'}_x=3x^2-63+12y$

$F^{'}_y=3y^2-63+12x$

$x^2-21+4y=0$

$y^2-21+4x=0$

2 Can we solve the system?

$4y=21-x^2$$(21-x^2)^2-21*16+4*16*x=0$

$441-42x^2+x^4-336+64x=0$

$105-42x^2+x^4+64x=0$

$x^4-42x^2+64x+105=0 \coloneqq P(x)$

$105=1*3*5*7$

$P(1)\neq0,~P(-1)=0,~P(3)=0,~P(-3)\neq0$

$P(5)=0,~P(-5)\neq0,~P(7)\neq0,~P(-7)=0$

Good, all roots (pairs of) are real (and symmetric).

$x=-7~~x=-1~~x=3~~~x=5$

$y=-7~~y=5~~~~~y=3~~~y=-1$

3 What does the second derivative test say?

$D=F^{''}_{xx}*F^{''}_{xx}-F^{''}_{xy}*F^{''}_{xy}$

$F^{''}_{xx}=6x$

$F^{''}_{yy}=6y$

$F^{''}_{xy}=12$

$D=36xy-144$

$D(-7, -7)>0,~~F^{''}_{xx} < 0 \implies \textnormal{local maximum}$

$D(-1, 5)<0~~~~~~~~~~~~~~~~~~~~\implies \textnormal{saddle point}$

$D(3, 3)>0,~~~~~~~~F^{''}_{xx}>0 \implies \textnormal{local minimum}$

$D(5, -1)<0~~~~~~~~~~~~~~~~~~~~\implies \textnormal{saddle point}$