multivariable-calculusoptimization
I'm kinda stuck on this one :
Find the minimum and maximum of the given function $f$ on $D$, where $$f(x, y) = xy$$ and $$D = \left\{(x,y) \in \mathbb{R}^2 : x^2+2y^2 \leq 1 \right\}$$
I don't know what to do with this domain $D$. I counted the first derivatives and got only point $(0,0)$ as a possible maximum/minimum inside $D$ but what about the boundary of $D$? Should I parametrize this ellipse or how should I approach this? Thanks for your tips!
Well, we have $f(x,y)=x^2y^2-2x-2y$ considered on the following green area:
So $f_x=2xy^2-2,~~f_y=2x^2y-2$ and solving $f_x=f_y=0$ gives us the critical point $x=y=1$ which is on the border of the area. Now think about the border: $$y=5,~~ 0\le x\le 5 \;\;\;\; x=0,~~0\le y\le 5 \;\;\;\; y=x,~~ 0\le x\le 5$$
If $y=5,~~ 0\le x\le 5$, so $$f(x,y)\to f(x)=25x^2-2x-10$$ and this easy for you to find the critical points which are maybe the best candidates on this border.
If $x=0,~~0\le y\le 5$, so $$f(x,y)\to f(y)=-2y$$ there is a similar way of finding such points on this border.
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
Obviously the function $f$ has no relative extreme inside the desired region $D$. In favt using the routine method we will find $(0,0)$ a saddle point in which $f_{xx}f_{yy}-(f_{xy}^2)<0$. Now consider $D$ and that $$x=\pm\sqrt{1-y^2}$$ Putting each parts $x=+$ and then $x=-$ separately we will find two one-variable functions $$f(y)=+y\sqrt{1-y^2}, ~~~f(y)=-y\sqrt{1-y^2}$$ I think you can find the relative extremes of these functions.... You have $4$ points as I plotted below: