Find the maximum and minimum value of a function constrained by an elipse

Question

Let $$f(x, y) = xy + y$$ Explain how we can determine that $f$ must attain both a maximum and a minimum value when $x, y$ varies along the part of the elliptical curve $x^2 + 3y^2 = 3$ that lies in the region where $ x \geq 0 $ and $ y \geq 0 $. Also, find the maximum and minimum values.

Solution:

The maximum and minimum can only be assumed at endpoints of the curve or points where Lagrange's conditions are satisfied (since singular points are absent). The endpoints are $(0, 1)$ and $(\sqrt{3}, 0)$. Here, $f(0, 1) = 1$ and $f(\sqrt{3}, 0) = 0$. To find points on the remaining part of the curve where a maximum and minimum can be assumed, we use Lagrange's conditions. Long story short, the only relevant point in the first quadrant is given by $x = 1$ and $y = \sqrt{\frac{2}{3}}$. There are no singular points. Then, $f(1, \sqrt{\frac{2}{3}}) = 2\sqrt{\frac{2}{3}} \approx 1.6$, and we see that this is the largest value. The smallest value is 0.

My question:

So I get the lagrange part, that is why I said "long story short", but what does the "endpoint of a close curve(elipse)" mean? How did they get the points $(0, 1)$ and $(\sqrt{3}, 0)$ from? I assume they let $g(x,y)= x^2 + 3y^2$ and did $\nabla g = (0,0)$. That is the only way I got those points but if I am right why are they doing that, if I am wrong what are they doing?

Answer 1

If you want to use the Lagrange multipliers technique you need to form a correct lagrangian first. Calling $f = x y + y$ we have

$$ L(x,y,\lambda,\mu,s) = f + \lambda(x^2+3y^2-3)+\mu_1(x-s_1^2)+\mu_2(y-s_2^2) $$

where $x-s_1^2=0$ and $y-s_2^2=0$ are equivalent to $x\ge 0$ and $y\ge 0$ respectively.

After that, the stationary points are the solutions for

$$ \nabla L = 0 = \left\{ \begin{array}{l} 2 \lambda x+\mu_1 +y \\ \mu_2 +6 \lambda y+x+1 \\ x^2+3 y^2-3 \\ x-s_1^2 \\ y-s_2^2 \\ \mu_1 s_1 \\ \mu_2 s_2 \\ \end{array} \right. $$

and after solving we have

$$ \left[ \begin{array}{cccccccc} f & x & y & \lambda & \mu_1 & \mu_2 & s_1^2& s_2^2\\ 0 & \sqrt{3} & 0 & 0 & 0 & -(1+\sqrt{3}) & \sqrt{3} & 0 \\ 1 & 0 & 1 & -\frac{1}{6} & -1 & 0 & 0 & 1 \\ 2 \sqrt{6}-4 \sqrt{\frac{2}{3}} & 1 & \sqrt{6}-2 \sqrt{\frac{2}{3}} & -\frac{1}{\sqrt{6}} & 0 & \sqrt{6} \left(\sqrt{6}-2 \sqrt{\frac{2}{3}}\right)-2 & 1 & \sqrt{6}-2 \sqrt{\frac{2}{3}} \\ \end{array} \right] $$

Note that when $s_k=0$ the corresponding restriction is actuating.

Answer 2

Let $$ x=\sqrt3\cos(\theta),y=\sin(\theta),\theta\in[0,\frac\pi2]. $$ Then \begin{eqnarray} f(x,y)&=&\sqrt3\cos(\theta)\sin(\theta)+\sin(\theta)\\ &=&2\sin(\theta)\sin(\theta+\frac\pi3)\\ &=&\cos(\frac\pi3)-\cos(2\theta+\frac\pi3)\\ &=&\frac12-\cos(2\theta+\frac\pi3). \end{eqnarray} Since $$ 0\le\theta\le\frac\pi2 $$ then $$ \frac\pi3\le2\theta+\frac\pi3\le\frac{4\pi}3 $$ and hence $$ -1\le\cos(2\theta+\frac\pi3)\le\frac{1}2 $$ So when $\theta=\frac\pi3$ or $x=\frac{\sqrt3}2,y=\frac{\sqrt3}2$, $f(x,y)$ reaches the max $\frac32$ and when $\theta=0$ or $x=\sqrt3,y=0$, $f(x,y)$ reaches the max $0$.