Mathematics: Calculus (Lagrange Multipliers)
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Mathematics: Calculus (Lagrange Multipliers)

[From: ] [author: ] [Date: 11-10-19] [Hit: ]
like (x,y) = (0, p/2) which has A = 0.I hope this helps!......
Use Lagrange multipliers to prove that the rectangle with maximum area that has a given perimeter p is a square.

10 points for the best explanation.

Thanks.

Z

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Let x and y be the dimensions of the sqaure.
So, we want to maximize A(x,y) = xy subject to the constant g(x,y) = 2x + 2y = p.

By Lagrange, grad A = λ * grad g
==> = λ<2, 2>.
==> y = 2λ and x = 2λ
==> y = x.

Hence, we have a square.

(Check: When x = y, 2x + 2x = p ==> x = p/4. So, A_max = p^2/16.
This is maximal, by checking any other point satisfying g, like (x,y) = (0, p/2) which has A = 0.)

I hope this helps!
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keywords: Lagrange,Mathematics,Calculus,Multipliers,Mathematics: Calculus (Lagrange Multipliers)
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