Find the absolute min and max values taken by the function $f(x,y)=2x^2+3y^2-4x-5$ on the disk $x^2+y^2\leq 16$

Question

This is what I currently have, not sure if the end of my solution is correct though.

$f_x=4x-4$ and $f_y=6y$

$4x-4=0$ therefore $x=1$

$6y=0$ therefore $y=0$

$f(1,0) = -7$ which is the critical point.

We now need to look at the boundary,

$$x^2+y^2=16$$

Rearranging, $$y^2=16-x^2$$

$$g(x)=-x^2-4x+43$$

We will need to find the absolute extrema of this function on the range $-4\leq x \leq 4$

$g'(x)=-2x-4$ implyng $x=-2$ when $-2x-4=0$

The value of this function at the critical point and the endpoints are,

$g(-4)=43$ $g(4)=11$ $g(1)=38$

We then need to find the y values which correspond

$x=-4: y^2=16-16=9$ so $x=0$

$x=4: y^2 = 16-16$ so $x=0$

$x=1: y^2=16-1=15$ so $y=\pm\sqrt{15}=\pm3.87$

The function values for $g(x)$ then correspond to the following function values for $f(x,y)$

...

So, comparing these values to the value of the function at the critical point of $f(x,y)$ that we found earlier we can see that the absolute minimum occurs at $(1,0)$ while the absolute maximum occurs at $(-4,0)$

Answer 1

$$f(x,y)\leq2x^2+3(16-x^2)-4x-5=-(x+2)^2+47\leq47.$$ The equality occurs for $x=-2$ and $y=\sqrt{12}$, which says that $47$ is a maximal value.

In another hand, $$f(x,y)\geq2x^2-4x-5=2(x-1)^2-7\geq-7.$$ The equality occurs for $x=1$ and $y=0$, which says that $-7$ is a minimal value.

Answer 2

You found the critical point of $g(x)$ to be $-2$ but then evaluated it at $1$ instead. If you evaluate it at $-2$ you get

$$g(-2)=47,$$

which is therefore the maximum (since $g(-2)>g(-4)$ and $g(-2)>g(4)$).

The corresponding $y$'s can, then, be found to be

$$ y = \pm\sqrt{16-(-2)^2}= \pm\sqrt{12}.$$

Hence, the maximum is $47$ and the maximands are $(x,y)=(-2,\sqrt{12})$ or $(x,y)=(-2,-\sqrt{12})$.