Find the critical points of [tex]f(x,y)[/tex]:
[tex]\dfrac{\partial f}{\partial x}=-2x=0\implies x=0[/tex]
[tex]\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12[/tex]
All three points lie within [tex]D[/tex], and [tex]f[/tex] takes on values of
[tex]\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}[/tex]
Now check for extrema on the boundary of [tex]D[/tex]. Convert to polar coordinates:
[tex]f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t[/tex]
Find the critical points of [tex]g(t)[/tex]:
[tex]\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0[/tex]
[tex]\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2[/tex]
[tex]\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi[/tex]
where [tex]n[/tex] is any integer. There are some redundant critical points, so we'll just consider [tex]0\le t< 2\pi[/tex], which gives
[tex]t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3[/tex]
which gives values of
[tex]\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}[/tex]
So altogether, [tex]f(x,y)[/tex] has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).
