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.

Wow, Lagrange multipliers in high school!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 findx = 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 = 8λλ=P/8so as expected we have a square with side length P/4: x=y=2λ=P/4