Question

Use LaGrange multipliers to prove that the rectangle with max 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 the perimeter P respectively. Find the Following. A = f(x,y) = xy; P = g(x,y) = 2x + 2y; grad f (x,y) = ?; lambda grad g = ? Then lambda = 1/2 y = x/2 implies that x = y; Therefore the rectangle with max area is a square with side length ? I have answered most of these questions, but there are three that I seem to be having trouble with. I thought grad f = , but lambda grad g should be a vector as well, neither one of them worked. Help with explaination would be very much appreciated. Thanks

Answer 1

Explanation:

The area function is f(x,y) = x*y the product of the values x and y

You want to maximize f under the constraint g(x) = p, with g(x) = 2x + 2y, the perimeter function. Lagrange Multiplier theorem states that, the point (x,y) where the area takes its maximun value should satisfy this equation:

grad(f) (x,y) = λ grad(g) (x,y), for certain λ.

In order to calculate the gradient of each function, you have to calculate the partial derivates fx, fy, gx, gy. The partial derivates can be obtained by deriving f and g assuming the other variable is a constant. For example, in order to obtain fx you assume that y plays like a constant, then

`f_x = (d)/(dx) xy = y` ;
`fy = (d)/(dy) xy = x` ;
`g_x = (d)/(dx) 2x+2y = 2` ;
`g_y = (d)/(dy) 2x+2y = 2`

Using this computations, we obtain

`grad(f) = (f_x,f_y) = (y,x)`

`\lambda grad(g) = (\lambda g_x,\lambda g_y) = (2 \lambda,2 \lambda)`

Therefore, we have that 2λ = y, 2λ = x

The first thing that you may realize is that x = y = 2λ. Since 2x + 2y = p, then 2x + 2y = 2x + 2x = 4x = p (because x = y). Therefore x = y = p/4, and 2λ = p/4, then λ = p/8.

We can conclude that:

• x = y = p/4 (the side lenght is p/4 for each side, so the rectangle is a sqaure)
• λ = p/8
• grad(f) = grad(g) = (y,x) = (2λ,2λ) = (p/8,p/8)
• area = x*y = p/4 * p/4 = p²/16

I hope i could helped you!