How Do You Solve Lagrange Multipliers for Circle Boundary Optimization Problems?

[thanksie037]

Homework Statement

Find the maximum and minimum values of f = (x-1)^2 + (y-1)^2 on the boundary of the circle g = x^2 + y^2 = 45.

Homework Equations

f=(x-1)^2 + (y-1)^2
g=x^2+y^2=45
gradf(x,y)=lambda*gradg(x,y)

The Attempt at a Solution

gradf(x,y)=<2x-2,2y-4>
gradg(x,y)=<2x,2y>
(1) 2x-2=lambda*2x
(2) 2y-4=lambda*2y
(3)g=x^2+y^2=45

solving the system for critical points:
x = 1/(1-lambda) plug into g?
y = 2/(1-lambda) plug into g?
gives lambda = 1 - (1/x) and lambda = 1 - (2/y)
set lambda = lambda:
1 - (1/x) = 1 - (2/y)
(1/x) = (2/y)
y = 2*x
here's where I get lost, how do i plus this back in? I shouldn't have to know the sqrt(45) to solve this. HELP

 

[gabbagabbahey]

Well, IF $y=2x$, then $$x^2+y^2=x^2+(2x)^2=5x^2=45 \Rightarrow x= \pm 3$$

But(!) you've made a mistake earlier in your calculations: Is $\vec{\nabla}f(x,y)$ really $\left< 2x-2, \, 2y-4 \right>$?

 

[thanksie037]

whoops f(x,y) = (x-1)^2+(y-2)^2
sorry about the typo

 

[thanksie037]

once I've solved to find that x=(plus or minus)3, where do I plug it back into find the critical points? do I plug it into (1) and solve for lamba? or do i plus it into (3) again and solve for y?

 

[gabbagabbahey]

well, you plug it into y=2x first, so that you can find the corresponding y values...then you plug your (x,y) pairs into your expression for f and find out which ones are maximums and which are minumums...for example, if I found that f(3,6)=2 and f(-3,-6)=4000 then I would know that the point (-3,-6) is a maximum and the point (3,6) was a minimum and that the maximum and minimum values of f were 4000 and 2 respectively.

 

[thanksie037]

do i need to find any more critical points? or will that suffice.. thank you for your help, by the way. your explanations have been clear, concise and beyond enlightening.

 

[gabbagabbahey]

Since all you are looking for are maximums and minimums, there are no other critical points to consider.