# lagrange multipliers

• Nov 9th 2009, 09:59 PM
calc626
lagrange multipliers
find the critical points of the function f(x,y)= ln(x^2+y^2) inside of the ring-shaped region R= (x,y): 4 <(or = to) x^2+y^2 >(or = to) 16

using lagrange multipliers find the critical points of f(x,y) on the first boundary and then on the second boundary

i got critical points x=0 y= +- sqrt(2) and y=0 x= +- sqrt (2) but i'm not sure

also i found the gradient of each and set them equal with a multiple of lambda so i eventually got..
x= +- sqrt((1-lamdba*y^2)/(lambda))
and y= +- sqrt((1-lamdba*x^2)/(lambda))

but im also not sure if this is right or where to go from there..

thanks for any help!
• Nov 10th 2009, 04:24 AM
HallsofIvy
Quote:

Originally Posted by calc626
find the critical points of the function f(x,y)= ln(x^2+y^2) inside of the ring-shaped region R= (x,y): 4 <(or = to) x^2+y^2 >(or = to) 16

using lagrange multipliers find the critical points of f(x,y) on the first boundary and then on the second boundary

i got critical points x=0 y= +- sqrt(2) and y=0 x= +- sqrt (2) but i'm not sure

also i found the gradient of each and set them equal with a multiple of lambda so i eventually got..
x= +- sqrt((1-lamdba*y^2)/(lambda))
and y= +- sqrt((1-lamdba*x^2)/(lambda))

but im also not sure if this is right or where to go from there..

thanks for any help!

What you say you have done is a bit confusing! Your object function is ln(x^2+ y^2). Its gradient is $\left(\frac{2x}{x^2+ y^2}\right)\vec{i}+ \frac{2y}{x^2+ y^2}\vec{j}\right)$ which is 0 only at (0,0) and that is not inside R.

On the inner ring, $4= x^2+ y^2$, you want to find critical points of $ln(x^2+ y^2)= ln(4)$. That is, the function is constant on that circle. Similarly, on the outer ring $x^2+ y^2= 16$ (you say "x^2+ y^2>= 16" but that can't be right. The ring is given by $4\le x^2+ y^2\le 16$) $ln(x^2+ y^2)= ln(16)$ a constant. Since there is no critical point inside the ring and the function is constant on both outer and inner boundaries, there are no "critical points". The minumum value of the function is ln(4) which it attains at every point on the inner boundary and the maximum function is ln(16) which it attains at every point on the outer boundary.

In fact, you could have made us of the symmetry by putting this in polar coordinates. The function would be $ln(r^2)$, $2\le r\le 4$ which is clearly an increasing function.
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