I am not sure if my method for this question is correct:
Given the function $$f(x,y)=x^2+y^2+2x+y$$, find it's minimum and maximum values about a closed disc of radius 2 centred at the origin.
I took the partial derivatives, $$f_x=2x+2; f_y=2y+1$$ and solved for $x$ and $y$. The minimum value is $f(0,0)$ and the maximum value is $f(-1,-1/2)$. Am I doing the correct thing? The answer I got is wrong according to the key.
[Math] Minimum and maximum values
multivariable-calculus
Related Solutions
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 $f_{\text{absolutemin}} = -4, f_{\text{absolutemax}} = 4$.
Lagrange multipliers do work for your problem, but that involves solving three simultaneous non-linear equations in three unknowns. There is another way that uses only one variable: parameterization.
Get a parameterization that describes the given curve in terms of only one variable. In the case of your ellipse you can use
$$x=\cos t, \quad y=\frac 17\sin t, \quad 0\le t\le 2\pi$$
Now you want to want to find the extrema of
$$\hat f(t)=4x+y=4\cos t+\frac 17\sin t$$
There are several ways to find the extrema of that, using calculus or just trigonometry. Here is a calculus way:
$$\hat f'(t)=-4\sin t+\frac 17\cos t=0$$ $$28\sin t=\cos t$$ $$\tan t=\frac 1{28}$$ $$\cos t=\sqrt{\frac 1{\tan^2 t+1}}=\pm\frac{28}{\sqrt{785}}$$ $$\sin t=\sqrt{1-\cos^2 t}=\pm\frac{1}{\sqrt{785}}$$ $$x=\cos t=\pm\frac{28}{\sqrt{785}}$$ $$y=\frac 17\sin t=\pm\frac{1}{7\sqrt{785}}$$
where $x$ and $y$ have the same sign. This means the maximum of $f$ is
$$4x+y=4\frac{28}{\sqrt{785}}+\frac{1}{7\sqrt{785}}=\frac{785}{7\sqrt{785}}=\frac{\sqrt{785}}{7}\approx 4.00255$$
and the minimum is the negative of that. This graph confirms the maximum.
Best Answer
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 $2y=x$.