# Homework Help: Help Please! Studying for test : Lagrange Multipliers

1. Dec 6, 2005

### love4math

Help Please! Studying for test : Lagrange Multipliers!!

Good morning all. I am having trouble with the next step to the following problem:

Q.Find all realtive extrema of x^2y^2 subject to the constraint 4x^2 + y^2 = 8.

g(x)= x^2y^2 f(x) = 4x^2 + y^2 = 8.

the gradiant of f = <8x,2y>
the gradiant of g = <2xy^2, 2x^2y>

therefore..

2xy^2 + 2x^2y = lamda (8x+2y)
lamda(8x) = 2xy^2 & lamda(2y)= 2x^2y
lamda = 2xy^2/8x & lamda = 2x^2y/2y
2x^2y/8x = 2xy^2/2y
4x^2 = y^2

I'm confused about how to find the extremas.
Can anyone help me?
Thanks all

2. Dec 6, 2005

### benorin

From

$$\vec{\nabla g}(x,y) =\lambda \vec{\nabla f}(x,y)$$

we know that $\left< 2xy^2, 2x^{2}y\right> = \lambda \left< 8x, 2y\right>$

cancel a 2, and get this system

$$xy^2 = \lambda 4x, x^{2}y = \lambda y$$

multiply the 1st by x, and the 2nd by y to get

$$x^{2}y^2 = \lambda 4x^2, x^{2}y^2 = \lambda y^2$$

so $\lambda 4x^2 = \lambda y^2$ so that if $\lambda \neq 0$, we have $4x^2 =y^2$.

then use the constraint $4x^2+ y^2 = 8 \Rightarrow y^2+ y^2 = 8$. Hence $y =\pm \sqrt{4} = \pm 2$.

From $4x^2 =y^2$, we have $x =\pm \sqrt{\frac {y^2}{4}} =\pm \frac {y}{2} = \pm 1$

Last edited: Dec 6, 2005