## anonymous 5 years ago Use Lagrange multipliers to find the maximum or minimum values of f(x,y) subject to the constraint. f(x,y)=3x-2y, x^2+2y^2=99

1. anonymous

Solution 1 (without multipliers): If you look at f you can see that its graph is a plane through the origin and grows with the highest rate in the direction of the gradient vector$\nabla f(x,y)=(3,-2)$ so you just have to go to the constraint (which is an origin centered circle with radius square root of 99) in that direction. This right point to the maximum is $\sqrt{99}(\frac{3}{\sqrt{13}},\frac{-2}{\sqrt{13}})$ where the value of f is $\sqrt{99}(\frac{9}{\sqrt{13}}+\frac{4}{\sqrt{13}})=\sqrt{99\cdot 13}\approx 35.87$ By symmetry you get the minimum is at point $\sqrt{99}(\frac{-3}{\sqrt{13}},\frac{2}{\sqrt{13}})$ where f is equal to -35.87. Solution 2 (using Lagrange multipliers):The function is $f(x,y,)=3x-2y$ and the constraint is $g(x,y)=x^2+y^2=99.$ Define the auxiliary function $\Lambda(x,y,\lambda)=f(x,y)+\lambda(g(x,y)-99)=3x-2y+\lambda(x^2+y^2-99).$ The system of linear equation formed by the partial derivatives of Lambda: $\begin{array}{rcl} \frac{\partial\Lambda}{\partial x}=3+2\lambda x&=&0\\ \frac{\partial\Lambda}{\partial y}=-2+2\lambda y&=&0\\ \frac{\partial\Lambda}{\partial \lambda}=x^2+y^2-99&=&0. \end{array}$ It has the solution: $x=-3\frac{\sqrt{99}}{\sqrt{13}},\quad y=2\frac{\sqrt{99}}{\sqrt{13}},\quad \lambda=\frac{1}{2}\frac{\sqrt{13}}{\sqrt{99}}$ and $x=3\frac{\sqrt{99}}{\sqrt{13}},\quad y=-2\frac{\sqrt{99}}{\sqrt{13}},\quad \lambda=-\frac{1}{2}\frac{\sqrt{13}}{\sqrt{99}}$ The first one gives the minimum and the second one the gives the maximum of the function subject to the given constraint.

2. anonymous

Oh... now i see it's 2y^2 at the constraint, so no big deal you should do almost the same but now solve the system of equation 3+2λx=0 −2+4λy=0 x^2+2y^2−99=0.