calc 3 - partial derivatives, lagrange multipliers

[Guest]

Hello all!

I was wondering if anyone could help me. I'm studying for a test on Tuesday and I can't figure this problem out - There are several of the same problems so if i saw how one was done I would deffinately pick up on the steps!

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


Thank you, hope someone can help!

Laura =)

 

[galactus]

\(\displaystyle F_{x}=2xy^{2}i\)

\(\displaystyle F_{y}=2yx^{2}j\)


\(\displaystyle {\nabla}f=2xy^{2}i+2yx^{2}j\)

\(\displaystyle {\nabla}g=8xi+2yj\)

So we have:

\(\displaystyle 2xy^{2}i+2yx^{2}j={\lambda}(8xi+2yj)\)

\(\displaystyle 2xy^{2}=8x{\lambda}\) and \(\displaystyle 2yx^{2}=2y{\lambda}\)

\(\displaystyle {\lambda}=\frac{2xy^{2}}{8x}\)

\(\displaystyle {\lambda}=\frac{2yx^{2}}{2y}\)

\(\displaystyle \frac{2xy^{2}}{8x}=\frac{2y^{2}}{2y}\)

\(\displaystyle 4x^{2}=y^{2}\)

Sub into the constraint equation and solve for x.

Be careful, there are various extrema to consider.

 

[Guest]

Hi, thank you for your help.

I just have one question

I'm confused how 4x^2 + y^2 = 8
Fx = 2xy^2
Fy = 2yx^2

when you are doing the partial derivative I thought when you are adding you do not consider the other term?

ex
4x^2 + y^2 - i thought this would equal 8x

which is different if it were 4x^2y^2 - i thought this equaled 8xy^2


I'm sorry there are many things to know and I get them confused at times!


Thanks again.
Laura