Answer
Absolute maximum: $f(1,0)=2$
Absolute minimum: $f(-1,0)=-2$
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, $f_x=6x^2,f_y=4y^3$
This yields, $x=0,y=0$ and $f(0,0)=0$
From the the given boundary $x^2+y^2=1$, we have $y=\sqrt {1-x^2}$
$f(x)=2x^3+(1-x^2)^2$ and $f'(x)=6x^2-4x(1-x^2)$
so, $x=-2, \dfrac{1}{2}$ and $y=\sqrt {1-x^2} =\dfrac{13}{16}$
Thus, Critical point: $(1,\sqrt 2)$
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 values at $2,-2$
Hence,
Absolute maximum: $f(1,0)=2$
Absolute minimum: $f(-1,0)=-2$
