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
10 points for the best explanation.
Thanks.
Z

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!
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
==>
==> 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!