Question:
find the maximum and minimum values of $ f(x,y)=4xy $ subject to $ y^2 + x^2 = 3 $. Use Use Lagrange multipliers λ and write this problem in the form $ L=f(x,y) + λg(x,y) $. Choose $g$ such that $g(0,0)=-3$.
(a) For which values of x do critical points of $f +λg$ occur?
(b) The max/min of $f(x,y)=4xy$ subject to $ y^2 + x^2 = 3 $ is:
for (a):
I have my Lagrange function $L=4xy + λ(y^2 + x^2 – 3)$ and the partial derivatives
$\delta L/\delta x = 4y+2 λ x$
$\delta L/\delta y = 4x+2 λ y$
$\delta L/\delta λ = y^2 + x^2 – 3x$
rearranging the 1st eqn and subbing into second gives $ x(4-\lambda^2)=0$ therefor $x=0$ or $\lambda = (+/-) 2$
I get stuck here and don't really know where to go after this? I think there should be two values of x that give critical points but im not sure what the other is or if $0$ is even right?
Best Answer
The system you have to solve is$$\left\{\begin{array}{l}\frac{\partial L}{\partial x}=0\\\frac{\partial L}{\partial y}=0\\x^2+y^2=3\end{array}\right.$$or$$\left\{\begin{array}{l}4y+2\lambda x=0\\4x+2\lambda y=0\\x^2+y^2=3\tag1\end{array}\right.$$The system which consists of the first two equations is equivalent to$$\left\{\begin{array}{l}\lambda x+2y=0\\2x+\lambda y=0,\end{array}\right.$$which is a system of two linear homogeneous equations, depending upon a parameter $\lambda$. If the matrix of coefficients of the system has determinant $0$, then its only solution is $(0,0)$, which is not a solution of the third equation of $(1)$. On the other hand, the matrix of coefficients of the system is $\left[\begin{smallmatrix}\lambda&2\\2&\lambda\end{smallmatrix}\right]$, whose determinant os $\lambda^2-4$, which is $0$ if and only $\lambda=\pm2$. So, its only necessary to solve the system $(1)$ in this case. When $\lambda=2$, its solutions are $\pm\left(\sqrt{\frac32},-\sqrt{\frac32}\right)$, and when $\lambda=-2$, its solutions are $\pm\left(\sqrt{\frac32},\sqrt{\frac32}\right)$.