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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

User Jyriand
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1 Answer

5 votes

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)

User Lidaranis
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7.7k points