Let $x,y$ and $z$ be positive real numbers . What is the maximum value of $xyz$ under the constraints $x^2+z^2=1$ and $y-x=0$ Please explain the process.
maximisation of value of a function under constraints
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optimization
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2You have $y=x$, so essentially you are maximizing $x^2z$ on a unit circle. Does that help ? – 2017-01-21
2 Answers
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From the last equation, we know that $x=y$. From the second equation we know that $x^2=1-z^2$. Thus we are interested in the maximum of
$$xyz=x^2z=(1-z^2)z$$
It's derivative set equal to $0$:
$$0=1-3z^2\implies3z^2=1\implies z^2=\frac13\implies z=\sqrt\frac13$$
Thus, the maximum value is
$$xyz\le\left(1-\frac13\right)\sqrt\frac13=\frac23\sqrt\frac13$$
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Alternatively, as mentioned by @Shailesh, you are maximizing a function over a unit circle.
More precisely, the intersection of $x^2+z^2=1$ and $y=x$ is the curve which can be parametrized as follows: $$ \begin{cases} x(t)=\cos t\\ y(t)=x(t)=\cos t\quad\quad \mbox{with }t\in [0,2\pi]\\ z(t)=\sin t \end{cases} $$ Therefore, the function that you are maximizing can be expressed as $$ x(t)y(t)z(t)=\cos^2(t)\sin(t), $$ which has maximum $\frac{2}{3\sqrt{3}}$.