Question

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

Answer 1

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.

Answer 2

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.