Find the extrema values of a multivariable function with one inequality constraint

Question

Question: Find the absolute maximum and minimum values of $f(x, y) = xy^2$ on the set $D=\{{(x, y), x^2+y^2 <= 3}\}$.

My attempted Solution: First I computed the following gradients of the functions:

Gradient of $f(x,y)=<y^2, 2xy>$

Gradient of $g(x,y)=<2x, 2y>$

Critical point 1: Since $<y^2, 2xy> = <0, 0>$, $(x, y) = (0, 0)$

By using the Lagrange multipliers, I obtained the following three functions:

$$y^2=2x\lambda$$ $$2xy=2y\lambda$$ $$x^2+y^2=3$$

However, I'm not sure how to continue on from this point to compute the critical points. Since $x$ and $y$ can be zero in both these cases, I'd assume that we can't divide either from both sides of the equation, thus making it hard to solve.

I'd be much appreciated if someone could show me how this is done. Thanks in advance!

Answer 1

$f(x,y) = xy^2$

With condition $x^2 + y^2 \leq 3$

Based on our function $(f(x,y) = xy^2)$, one thing to note is that our extrema would likely occur on the boundary $x^2 + y^2 = 3$.

Continuing from the point that you got to in your question,

$y^2=2x\lambda \implies \lambda = \frac{y^2}{2x} \, $ if $x \ne 0$

$2xy=2y\lambda \implies \lambda = x \,$ or $ \,x = 0$

For $x = 0, y = 0$

For $x \ne 0$, equating $\lambda, \, y = \pm \sqrt2 \, x$

Based on the condition, $-1 \leq x \leq 1$

So potential points to check for min and max are $(-1, - \sqrt2), (-1, \sqrt 2), (1, -\sqrt2), (1, \sqrt 2), (0, 0)$.

Answer 2

An alternative approach with change of variables. Let $x=r\cos{\theta}$ and $y=r\sin{\theta}$.

Then we have as our inequality $r^2\leq 3$ and we seek to find the extreme values for $f(r,\theta)=r^3 \cos{\theta} \sin^2{\theta}$.

From here we can use single variable calculus to find max and min values of $f$. Max and min for $r^3$ are 27 and 0, respectively.

Let $ g(\theta)=\cos{\theta}\sin^2{\theta}=\cos{\theta}-\cos^3{\theta}$.

$g'(\theta)=-\sin{\theta}+3\sin{\theta}\cos^2{\theta}$. It's zeros occur when theta is an integer multiple of $\pi$ or $\cos^2{\theta}=1/3$. $g$ is zero at integer multiples of pi and takes on extremes of $\frac{2}{3\sqrt{3}}$ and $\frac{-2}{3\sqrt{3}}$. $g$ flips signs by adding $\pi$ to the argument.

Altogether, $f$ has the extreme values $6\sqrt{3}$ and $-6\sqrt{3}$.