Find the absolute maximum and minimum values of f on the set d. f(x, y) = xy2 7, d = (x, y)

Question

Find the absolute maximum and minimum values of f on the set d. f(x, y) = xy2 7, d = x ≥ 0, y ≥ 0, x2 y2 ≤ 3

Answer 1

Assuming you mean
`f(x,y) = xy^2` over the domain

`D = \left\{(x,y) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } x^2 + y^2 \le 3\right\}`

we first observe that
`f(x,y) = 0` for all
`(x,y)` on the coordinate axes.

There are no critical points elsewhere in the interior of
`D`, since

`(\partial f)/(\partial x) = y^2 = 0 \implies y=0`

`(\partial f)/(\partial y) = 2xy = 0 \implies x = 0 \text{ or } y = 0`

Parameterize the circular arc boundary by
`x=\sqrt3\cos(t)` and
`y=\sqrt3\sin(t)`, where
`0\le t\le\frac\pi2`. Then

`f(x(t), y(t)) = g(t) = 3\sqrt3 \cos(t) \sin^2(t) = 3\sqrt 3 (\cos(t) - \cos^3(t))`

Find the critical points of
`g`.

`g'(t) = -3\sqrt3 \sin(t) + 9\sqrt3 \cos^2(t) \sin(t) = 0`

`-3 \sin(t) (1 - 3 \cos^2(t)) = 0`

`\sin(t) = 0 \text{ or } 1 - 3 \cos^2(t) = 0`

`\sin(t) = 0 \text{ or } \cos^2(t) = \frac13`

`\sin(t) = 0 \text{ or } \cos(t) = \pm\frac1{\sqrt3}`

In the first case, we get

`t = \sin^(-1)(0) + 2n\pi \text{ or } t = \pi - \sin^(-1)(0) + 2n\pi`

where
`n` is an integer; the only solution on the boundary of
`D` is
`t=0` corresponding to the point
`(\sqrt3,0)`.

In the second case, we get

`t = \cos^(-1)\left(\frac1{\sqrt3}\right) + 2n\pi \text{ or } t = -\cos^(-1)\left(\frac1{\sqrt3}\right) + 2n\pi`

with only one relevant solution at
`t=\cos^(-1)\left(\frac1{\sqrt3}\right)` corresponding to
`(1,\sqrt2)`.

In the third case, we get

`t = \cos^(-1)\left(-\frac1{\sqrt3}\right) + 2n\pi \text{ or } t = -\cos^(-1)\left(\frac1{\sqrt3}\right) + 2n\pi`

but there is no
`t` in this family of solutions such that
`0\le t\le\frac\pi2`.

So, we find

`\min\left\{xy^2 \mid (x,y) \in D\right\} = 0 \text{ at } (0,0)`

(but really any point on either axis works)

`\max \left\{xy^2 \mid (x,y) \in D\right\} = 2 \text{ at } (1,\sqrt2)`