Resolve this system:

Question

Im tried to resolve this problem:

$$\max\quad f\left( x,y \right) =xy\quad \text{s.a}\quad \begin{cases} x^2 +y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$

Well, i form the lagrangian and the respective gradient, so i had this system to resolve:

$$\begin{cases} y+2\lambda_1 x+\lambda_2=0 \\ x+2\lambda_1 y+ \lambda_2=0 \\ x^2+y^2+z^2 -1=0 \\ x+y+z=0 \end{cases}$$

And i can't find all the solutions, i need help-

Answer 1

W/O using calculus,

Let $x=\cos A\cos B,y=\cos A\sin B,z=\sin A$

$\implies \cos A\cos B+\cos A\sin B+\sin A=0\ \ \ \ (1)$

Method $\#1:$ $\iff\dfrac{-\sin A}{\cos B+\sin B}=\dfrac{\cos A}1=\pm\sqrt{\dfrac1{(\cos B+\sin B)^2+1}}$

$\implies\cos^2A=\dfrac1{2+2\cos B\sin B}$

Method $\#2:$

By $(1),\tan A=-(\cos A+\cos B)$

$\cos^2A=\dfrac1{1+\tan^2B}=\dfrac1{1+(\cos A+\cos B)^2}=\dfrac1{2+2\cos B\sin B}$

By anyone of the methods, we need to maximize $xy=\cos^2A\cos B\sin B=\dfrac{\cos B\sin B}{2+2\cos B\sin B}$

$2xy=\dfrac{\sin2B}{2+\sin2B}=1-\dfrac2{2+\sin2B}$

We need to maximize $\sin2B$ which is $\le1$

Answer 2

$z=-(x+y), x^2+y^2+(x+y)^2=1 \implies x^2+y^2+xy=\dfrac{1}{2},x^2+y^2\ge 2xy \implies 3xy \le \dfrac{1}{2} \iff xy\le \dfrac{1}{6}$, when $x=y \cap x^2=\dfrac{1}{6}$