Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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 .

Sagot :

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

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.