Given the area of a rectangle is A = LW. If three sides of the rectangle are fixed, I.e. 2W + L = 40, find L and W such that you maximize the area of the rectangle.
Is it L=10, W=20?
I meant W=10, L=20
2 W + L = 40
so
L = 40 - 2 W
A = W L = W (40 - 2 W) = 40 W - 2 W^2
dA/dW = 40 - 4 W
W = 10
so yes
To maximize the area of a rectangle given a constraint, we can use calculus. Let's begin by solving the constraint equation 2W + L = 40 for L:
L = 40 - 2W
Now substitute this expression for L into the area formula A = LW:
A = (40 - 2W)W
To find the maximum area, we need to find the critical points of A, which occur when its derivative dA/dW is equal to zero. Let's find this derivative:
dA/dW = 40 - 4W
Set dA/dW to zero and solve for W:
40 - 4W = 0
4W = 40
W = 10
Now substitute this value of W back into the equation 2W + L = 40 to find L:
2(10) + L = 40
20 + L = 40
L = 20
Therefore, the values that maximize the area of the rectangle are L = 20 and W = 10. These values result in a maximum area because the derivative dA/dW changes from positive to negative at this point, indicating a maximum.