Find the absolute maximum and minimum values of f on the setd. f(x, y) = xy2 + 2, d = {(x, y) | x ≥ 0, y ≥ 0, x2 + y2 ≤ 3}

[tex]f(x,y)=xy^2+2[/tex]
[tex]\implies\begin{cases}f_x=y^2=0\\f_y=2xy=0\end{cases}[/tex]

which occurs whenever [tex]y=0[/tex] or [tex]x=0[/tex]. Under either condition, we have [tex]f(x,0)=f(0,y)=2[/tex]. The equations above admit no other critical points on the surface, so we consider the boundaries. We already have [tex]x=0[/tex] and [tex]y=0[/tex] accounted for, so let's look at the circular part.

We can parameterize this by

[tex]\begin{cases}x=\sqrt3\cos t\\y=\sqrt3\sin t\end{cases}[/tex]

with [tex]0\le t\le\dfrac\pi2[/tex]. So now

[tex]f(x,y)=f(x(t),y(t))=F(t)=(\sqrt3\cos t)(\sqrt3\sin t)^2+2[/tex]
[tex]\implies F(t)=3^{3/2}\cos t\sin^2t+2[/tex]

which can be written in terms of cosine alone as

[tex]\implies F(t)=3^{3/2}(\cos t-\cos^3t)+2[/tex]

Now, we get critical points at

[tex]F'(t)=3^{3/2}\sin t(3\cos^2t-1)=0\implies t=0\text{ or }\arccos\dfrac1{\sqrt3}[/tex]

We omit [tex]t=0[/tex], because that gives us [tex]F(0)=2[/tex] again. Meanwhile, when [tex]t=\arccos\dfrac1{\sqrt3}[/tex], we get [tex]F\left(\arccos\dfrac1{\sqrt3}\right)=4[/tex].

Finally, we compute the corresponding critical point in terms of [tex]x,y[/tex]. We have

[tex]\begin{cases}x=\sqrt3\cos\left(\arccos\dfrac1{\sqrt3}\right)=1\\\\y=\sqrt3\sin\left(\arccos\dfrac1{\sqrt3}\right)=\sqrt2\end{cases}[/tex]

So to summarize, minimum values of 2 are obtained anywhere along the borders where [tex]x=0[/tex] or [tex]y=0[/tex], and a maximum value of 4 at [tex](x,y)=(1,\sqrt2)[/tex].