My Math Forum Direction of max increase

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 June 7th, 2010, 07:13 PM #1 Newbie   Joined: Jun 2010 Posts: 6 Thanks: 0 Direction of max increase Find the direction of max increase of$f(x,y)=6x^2+3y^2$ at point (1,-1) I got 12i-6j ? Also how do you find the max constraint of f(x,y)=4xy subject to the constraint x^2+ (y+1)^2 = 1 ? Thanks
 June 8th, 2010, 06:45 AM #2 Newbie   Joined: Jun 2010 Posts: 6 Thanks: 0 Re: Direction of max increase Am I on the wrong forum?
June 8th, 2010, 08:37 AM   #3
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Re: Direction of max increase

Quote:
 Originally Posted by jj007 Am I on the wrong forum?
Right forum.
The max increase result is correct.
Have you learned Lagrange multipliers?

 June 8th, 2010, 08:57 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Direction of max increase $x^2+(y+1)^2=1$ if and only if there exists $\theta\in[0,2\pi)$ such that $x=\cos\theta,\ y=\sin\theta-1.$ Therefore $f(x,y)=4\cos\theta(\sin\theta-1)$ for some $\theta\in[0,2\pi)$ iff $x^2+(y+1)^2=1.$ Differentiate the expression $4\cos\theta(\sin\theta-1)$ to find its extrema, and then figure out which is the maximum.

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