Given the function $f(x,y)=x^4+y^4$.
while $x$ and $y$ must be on the circle $x^2+y^2=16$
Find the maximum and minimum points of the function $f(x,y)$.
I have used Lagrange multipliers setting lagrange function $\phi(x,y,\lambda)=x^4+y^4+\lambda(x^2+y^2-16)$, and got final answer $x= \pm \sqrt{8}$, $y= \pm \sqrt{8}$.
And I was confused if I have found two points $((\sqrt{8} , \sqrt{8}), (- \sqrt{8}, - \sqrt{8}))$ or four points (all possible values), and wanted to make sure my solution is correct.
So to test myself, I have decided to use another method and got stuck also, here's what I tried:
Let $x=4\cos(t)$, $y=4\sin(t)$ while $2\pi \ge t \ge 0$.
and I substitute that into the $f(x,y)$ to make sure the points I will find are on the circle.
$u(t) = 256\cos(t)^4 + 256\sin(t)^4$. to find maximum and minimum I took derivative:
$u'(t) = 1024 ( -\cos(t)^3\sin(t) + \sin(t)^3\cos(t) = 0$
$512 ( -\cos(t)^2\sin(2t) + \sin(t)^2\sin(2t)=0$
$512\sin(2t) (\sin(t)^2-\cos(t)^2) = 0$
So if $\sin(2t) = 0$ I got that $t = 0, \frac{\pi}{2},2\pi$
and here I got alot of different points that I don't know which of them are the true values and I feel lost in my calculations.
I would love to hear feedback about my second approach, and to make sure if my answer in the first approach is true or not.
Any help is really appreciated, thanks in advance!
You need to combine all possibilities of signs of your solutions for $x$ and $y$.
In your second approach you forgot some solutions, namely $\pi$, $\frac{3\pi}{2}$ and the solutions $0$ and $2\pi$ of course are the same since the resulting points $(4\cos(x),4\sin(x))$ coincide.
Even further, the $sin^2-cos^2$ term also has some zeros. While these zeros correspond to what you've found in your first approach, the points you found in your second approach are missing in your first approach (Hint: In general, an equation $x^3-x=0$ has not only the solutions $x=\pm1$ but also the solution $0$.)