Use Lagrange to find absolute max/min values

[MarkSA]

Hi,

Use Lagrange Multiplier method to find the absolute max and min values of f(x,y) = xy subject to the constraint that x[sup:2hd5qpt8]2[/sup:2hd5qpt8] + y[sup:2hd5qpt8]2[/sup:2hd5qpt8] = 18.

I let g(x,y) = x[sup:2hd5qpt8]2[/sup:2hd5qpt8] + y[sup:2hd5qpt8]2[/sup:2hd5qpt8] - 18
f[sub:2hd5qpt8]x[/sub:2hd5qpt8] = y
f[sub:2hd5qpt8]y[/sub:2hd5qpt8] = x
g[sub:2hd5qpt8]x[/sub:2hd5qpt8] = 2x
g[sub:2hd5qpt8]y[/sub:2hd5qpt8] = 2y

k = lagrange multiplier
Then setting the parts equal from above, y = 2xk, and x = 2yk.
Solve the system of equations for k i get: k = +/- 1/2.
So here is my problem. When I go to plug k back into y=2xk or x=2yk, I get y=x or y=-x, which seems to suggest that the only critical point is (0,0). I know that's not right though. Where am I going wrong?

 

[galactus]

\(\displaystyle yi+xj={\lambda}(2xi+2yj)\)

\(\displaystyle y=2x{\lambda}, \;\ x=2y{\lambda}\)

\(\displaystyle {\lambda}=\frac{y}{2x}, \;\ {\lambda}=\frac{x}{2y}\)

\(\displaystyle \frac{y}{2x}=\frac{x}{2y}\)

\(\displaystyle y=-x, \;\ y=x\)

Subbing into the constraint, we get:

\(\displaystyle 2x^{2}=18\Rightarrow x=\pm 3\)

 

[MarkSA]

thanks, that was it. I'd forgotten that it has to be subbed back into the constraint