# Lagrante multipliers check + general question

• Jan 6th 2013, 04:12 PM
jones123
Lagrante multipliers check + general question
Hello! First of all I'd like somebody to check if what I'm doing is correct. Second question: see end

I need to find the extrema of 6 - 4x - 3y subject to x² + y² = 1.

∂f/∂x = -4 and ∂g/∂x = 2x and ∂f/∂y = -3 and ∂g/∂y = 2y

So we've got:
{ -4 = 2
λx
{ -3 = 2λy
{ x² + y² = 1

<=>

{ -2/x =
λ
{ -3/2y = λ
{ x² + y² = 1

If -2/x = -3/2y then 3x = 4y, this means 3x/4 = y

Substitute this into contraint:
x² + (3x/4)² = 1 <=> 25x² - 16 = 0 <=> x =
±4/5

If y = 3x/4, then 3(
±4/5) / 4 = ±3/5

Now here comes my second question. Now that I've got
x = ±4/5 and y = ±3/5, how do I know that the points are
(4/5, 3/5) and (-4/5, -3/5)
OR
(4/5, 3/5) , (-4/5; -3/5) , (4/5; -3/5) , (-4/5, 3/5) ?

Thanks!
• Jan 6th 2013, 04:53 PM
russo
Re: Lagrante multipliers check + general question
Hi, I've redone the exercise and yes the critical points are right.

Regarding to your second question, as I've done it before, I think you use the points wich satisfies your $\displaystyle \lambda$ identity ($\displaystyle y = \frac{3}{4} x$), so if $\displaystyle x=\frac{4}{5}$ then $\displaystyle y=\frac{3}{5}$ and so on. So you actually check the first pair. I think is that way but still if you can check it, better.
• Jan 6th 2013, 06:32 PM
MarkFL
Re: Lagrante multipliers check + general question
Yes, you have only two critical values for the reason given by russo.