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) = ????∇g = Then ???? = 1 2 y = implies that x = . Therefore, the rectangle with maximum area is a square with side length .

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jtellezd jtellezd · Answer 1 · 2021-04-04T15:59:17+02:00

Answer:

a) Rectangle of maximum area ( given perimeter p ) is

A= x² That means the rectangle of maximum area, is a square

Step-by-step explanation:

The equation: A (x,y) = x*y is the area of a rectangle ( to maximize)

Subject to: p = 2*x + 2*y or g(x,y) = 2*x + 2*y -p

Now

A(x,y) = x*y δA/δx = y δA/δy = x

And

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

δg(x,y)/δx = 2 and δg(x,y)/δy = 2

Matching respective partial derivatives we get a system of equation

δA/δx = y = λ * = δg(x,y)/δx = 2

y = 2*λ

δA/δy = x = 2*λ

The system of equations is:

y = 2*λ

x = 2*λ

And 2*x + 2*y -p = 0

p = 2*x +2*y

So x = y p is equal either 4*x or 4*y

Solving for λ