Question

#LagrangeMultipliers Find the stationary points of the function f(x,y) = xy subject to the
constraint 1/x + 1/y = 1, and find the value for f at these points.
So far I have,

\[f(x,y) = xy\]\[g(x,y) = {1 \over x} + {1 \over y} -1\]
\[H = f(x,y) - \lambda[g(x,y)]\]\[H = xy - {\lambda \over x} - {\lambda \over y} + \lambda \]
\[{\delta H \over \delta x} = {y +{ \lambda \over x^2}} = 0\]\[{\delta H \over \delta y} = {x +{ \lambda \over y^2}} = 0\]\[{\delta H \over \delta \lambda} = 1 - {1 \over x} - {1 \over y} = 0\]

Need help to finish off please!

Answer 1

@Directrix @imranmeah91@Hero @LagrangeSon678 @Mertsj

Answer 2

I would try to help, but this looks like such a different way than how I do it...

Answer 3

\[\nabla f=\lambda\nabla g\]\[\langle y,x\rangle=\lambda\langle-\frac1{x^2},-\frac1{y^2}\rangle\]\[-x^2y=\lambda=-y^2x\]something like that if I remember correctly...

Answer 4

From \(\Large\frac{\partial{H}}{\partial{x}}=0\), we have \(\Large \lambda=-x^2y\). Substitute this into \(x+\frac{\lambda}{y^2}=0\) then solve for x and y.

Answer 5

So I get x = y, what can i do with that?

Answer 6

turingtests method is the MIT way as well, alot simpler since I have tried both ways and like that alot better

Answer 7

id have to review it myself to be sure of a solution

http://www.academicearth.org/lectures/lagrange-multipliers

Answer 8

You solve your 3 equations and you get
\[
\lambda = -8,\, x = 2,\, y =2
\]

Answer 9

You also can transform it into a function of one variable by
replacing
\[ x=\frac{y}{y-1} \\

\]
to get
\[
g(y)= \frac{y^2}{y-1} \\
g'(y)=\frac{y^2-2 y}{(y-1)^2}\\

\]
You can see that g'=0 at y=0 and y =2.
We reject y=0 Why?
if y=2, then x=2