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 .

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

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: