Use lagrange multipliers to prove that the rectangle with maximum area that has a given perimeter p is a square. let the sides of the rectangle be x and y and let f and g represent the area (a) and perimeter (p), respectively. find the following. a = f(x, y) = p = g(x, y) = f(x, y) = lambda g = then lambda = 1/2y = implies that x = therefore, the rectangle with maximum area is a square with side length.

the rectangle of maximum area is a square with side length x=y= p/4

Step-by-step explanation:

Since

A = f(x,y) = x*y

p= g(x,y)= 2*x+2*y

therefore using Lagrange multiplier →λ

F(x,y) = f(x,y) - λ*g(x,y)

and

Fx(x,y) = fx (x,y) - λ*gx(x,y) = 0

Fx(x,y) = fy (x,y) - λ*gy(x,y) = 0

g(x,y) = p

where fx and gx are the partial derivatives with respect to x and fy and gy are the ones of y

therefore

fx (x,y) - λ*gx(x,y) = y - λ*2 = 0 → y = 2λ

fy (x,y) - λ*gy(x,y) = x - λ*2 = 0 → x = 2λ

2*x+2*y = p → 2*2λ + 2*2λ = p → 8λ = p →  p/8=λ

therefore

x = y = 2λ = 2*p/8 = p/4

A max = x*y = p/4 * p/4 = p²/16

since the side x = side y → the rectangle of maximum area is a square with side length = p/4

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step-by-step explanation:

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step-by-step explanation: