Finding local maxima and minima of function

Question

I am supposed to find the local extreme of function $z=xy$ on set $x^2+y^2=1$.

I used substitution:$ x=\cos \theta ,y=\sin \theta $, where $\theta \in \left [ 0,2\pi \right ]$.

So:

$z=\cos \theta \sin \theta \iff$ $z=\frac{\sin 2\theta }{2}$, but I do not know, how to continue.

Can anyone help me, please?

Answer 1

Let $\phi=2\theta$, so we're looking for the extremes of $z=\dfrac{\sin2\theta}2=\dfrac{\sin\phi}2.$

Since $-1\le \sin\phi\le 1$, the extremes occur when $\sin\phi=\sin2\theta=\pm1$; i.e., $z=\pm\dfrac12$.

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Another approach would be to look for extremes of $z^2=x^2(1-x^2)=x^2-x^4;$

the derivative is zero when $2x-4x^3=2x(1-2x^2)=0$.

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Another approach uses a Lagrange multiplier:

$xy-\lambda(x^2+y^2-1)$

$y-2\lambda x=0, x-2\lambda y=0, x^2+y^2=1$

$y=2\lambda x=4\lambda^2y$ so $\lambda^2=\frac14$ so $\lambda=\pm\frac12$ so $y=\pm x$ and $x^2+y^2=1...$

Answer 2

$$1 = x^2 + y^2 \ge 2 |x|\cdot|y| $$

$$ \rightarrow 1 \ge 2|z| $$

$$ \rightarrow \frac{1}{2} \ge z \ge -\frac{1}{2} $$

$ \rightarrow $ local extreme of $z$ is $ \pm\frac{1}{2} $

Answer 3

The sine function reaches its maximum at $\pi/2\pm2n\pi,$ so you'd have $$ 2\theta = \frac \pi 2 \pm 2n\pi $$

$$ \theta = \frac \pi 4 \pm n\pi $$

This corresponds to two points on the circle: \begin{align} & \theta = \frac\pi4, \text{ so that } (x,y) = \left( \frac{\sqrt 2} 2, \frac{\sqrt 2} 2 \right) \\[10pt] & \theta = \frac \pi 4 + \pi, \text{ so that } (x,y) = \left( \frac{-\sqrt 2} 2, \frac{-\sqrt 2} 2 \right) \end{align}