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.
(i)- A = f(x, y)
(ii)- p = g(x, y)
(iii)- ∇f(x, y)
(iv)- γ∇g
(v)-Then γ=1/2y=_ Implies that x=_
(vi)-Find the rectangle with maximum area is a square with side length.

As a rule with these Lagrange multiplier problems, when the problem is symmetrical with respect to interchange of the variables, the solution almost always ends up with all the variables equal -- what else could it be?

We want to maximize the area of a rectangle with sides x and y subject to the perimeter being constant.

(i)

The area of a rectangle is just the product of its sides:

A = f(x,y) = xy

(ii)

The perimeter of a rectangle is the sum of its sides:

P = g(x,y) = x + x + y + y = 2x+2y

(iii)

Usually I like to form the objective function E=f-λg before I take the derivatives. I usually use a lambda not a gamma for the multiplier.

Let's do what they ask. They want the gradient ∇f(x, y)

[tex]f(x,y)=xy[/tex]

[tex]\dfrac{\partial f}{\partial x} = y[/tex]

[tex]\dfrac{\partial f}{\partial y} = z[/tex]

∇f(x, y) = (y, x)

(iv)

[tex]g(x,y) = 2x+2y[/tex]

[tex]\dfrac{\partial G}{\partial x} = 2[/tex]

[tex]\dfrac{\partial G}{\partial y} = 2[/tex]

λ∇g(x, y) = (2λ, 2λ)

(v)

I'm not sure what γ=1/2y is about; I'll solve it like I know how and see where we are.

[tex]E = f - \lambda g = xy - 2\lambda(x+y)[/tex]

[tex]0=\dfrac{\partial E}{\partial x} = y -2\lambda[/tex]

There it is. We get

y = 2λ

[tex]0=\dfrac{\partial E}{\partial y} = x -2\lambda[/tex]

so we also find

x = 2λ

(vi)

We have y=x=2λ so we've shown the variables are equal, i.e. our rectangle is a square. We can solve for λ using our constraint:

P = 2x+2y = 8λ

λ=P/8

so as expected we have a square with side length P/4:

x=y=2λ=P/4