multivariable-calculus
I need to find absolute maximum and minimum values of
$$f(x,y)=4x^3+3y^2$$
on a unit disk $D={(x,y)|x^2+y^2\le 1}$
I thought about finding $f_x$ and $f_y$ first to find in the critical point is in $D$ or not. I think that it is because I found it to be $(0,0)$. From here we have to look at the boundary, am I correct? How do I proceed?
If you have a boundary which is given by the level curve of a function (here $x^{2} + y^{2}$), you can use the Lagrange multiplier theorem (see here). You have to check the points of the boundary such that the gradient of the function you want to study and the gradient of the function giving the domain are parallel.
In this case you have that the only critical point in $\mathbb{R}^{2}$ is $\left(-1 , -1/2 \right)$. $f$ is the sum of two quadratic functions in $x$ adn $y$ separately and with coefficients of $x^{2}$ and $y^{2}$ both positive. Hence $\mathrm{lim}_{||\left(x,y\right)|| \to +\infty}f\left(x,y\right)=+\infty$, so the point we have is the absolute minimum, and it lies in the circle. Now to find the maximum in the circle check the points of the circle of radius 2 such that $\left(2x+2,2y+1\right)$ is parallel to $\left(2x,2y\right)$. If you put them in a matrix and want the determinant to be $0$, it gives you the condition $2y=x$.
I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.
edit 1:(add more details)
$f(x,y)=(x-2)^2+y^2+5 \implies (x-2)^2+y^2=f(x,y)-5=R^2$
you want to find max and min of $f(x,y)$ is same to find max and min of $R$.
note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$
if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.
Best Answer
Let $x = r\cos \theta, y = r\sin \theta \to f(r,\theta) = 4r^3\cos^3 \theta + 3r^2\sin^2 \theta=4r^3\cos^3\theta + 3r^2(1-\cos^2\theta)=3r^2+4r^3\cos^3\theta-3r^2\cos^2\theta$. Let $t = \cos \theta \to f(r,\theta) = f(r,t) = 3r^2+4r^3t^3-3r^2t^2$ with $0 \leq r \leq 1, -1 \leq t \leq 1$. Now taking partials we have: $f_r = 6r(1+2rt^3-t^2) = 0 = f_t = 6r^2t(2rt-1)$. Now if $rt \neq 0$, then: $1+2rt^3-t^2 = 0 = 2rt-1 \to 0=1 + t^2(2rt-1) = 1 + 0 = 1$, contradiction. Thus $rt = 0$. If $r = 0 = t \to x = 0 = y$. If $r = 0$ only ,then $x = r\cos \theta = 0 = r\sin \theta = y$. If $t = 0$ only, then $x = rt = 0, y = \pm r$. From this we conclude that: $f_{\text{localmin}} = 0, f_{\text{localmax}} = 3$, and considering the endpoints we evaluate $f(r,t)$ at $(r,t) = (0,\pm1),(1,\pm1)$, and have $f_{\text{absolutemin}} = -4, f_{\text{absolutemax}} = 4$.