Answer
Absolute maximum: $f(2,2)=f(-1,2)=18$
Absolute minimum: $f(-2,-2)=-18$
Work Step by Step
To calculate the critical points we have to first put $f_x(x,y), f_y(x,y)$ equal to $0$.
Thus, $f_x=3x^2-3,f_y=-3y^2+12$
This yields, $x=\pm 1,y=\pm 2$
Thus, Critical point: $(1,2),(-1,2)$
At $(1,2):$f(1,2)=14$ and At $(-1,2):$f(-1,2)=18$
From the the vertices of a quadrilateral, we have
so, $x=-2, \dfrac{1}{2}$ and $y=\sqrt {1-x^2} =\dfrac{13}{16}$
At $(x,y)=(1,\sqrt 2)$
$f(1,\sqrt 2)=2$
Now, $f(1,0)=2x^3+y^4=2$
and $f(-1,0)=2x^3+y^4=-2$
Therefore, it has minimum and maximum value are at $2,-2$
Hence,
Absolute maximum: $f(2,2)=f(-1,2)=18$
Absolute minimum: $f(-2,-2)=-18$
