Question

Use la grange multipliers to find the indicated extrema, assuming that x and y are positive.

f(x,y) = x^2-y^2
constraint = g(x,y): 2y-x^2=0

Answer 1

Partial derivatives of f and g
\[\Delta f(x,y)= <2x,-2y>\]
\[\lambda \Delta g(x,y)=<-2\lambda x,2\lambda>\]

Answer 2

\[2x=-2 lambda x\]
\[-2y=2\lambda\]

Answer 3

x would be \[\lambda x\] and y \[2\lambda\] right?

Answer 4

oh wait I mean y would be just lambda

Answer 5

\[\nabla f=\lambda \nabla g\]\[\left(\begin{matrix}2x \\ -2y\end{matrix}\right)=\lambda \left(\begin{matrix}-2x \\ 2\end{matrix}\right)\]\[2x=-2x \lambda\]simplifying gives lambda=-1\[-2y=2 \lambda\] simplifying gives y=-lambda, y=1
from your constraint\[2y-x ^{2}=0\]\[x ^{2}=2\]\[x=\pm \sqrt{2}\]

Answer 6

Oh okay I get it now Thank you.