Question

For a rectangle with sides L and D, the perimeter 2(L+D) is constrained to be a constant C. If L=0,D=C/2 and the area is zero. Id D=0, L=C/2 and the area is zero. Vary L over the range 0 to C/2. Plot the area as a function of L and determine the optimum values of L and D that maximize the area A. Assume C=200 meters.

Answer 1

The optimum values for L and D that maximize the area A of the rectangle, given the constraints, are L = C/4 and D = C/4. The reason for this is that it ensures a balanced distribution of the constant perimeter, resulting in the maximum area.

Step-by-step explanation:

In this problem, we are tasked with maximizing the area A of a rectangle with sides L and D, subject to the constraints of a constant perimeter, given by 2(L + D) = C, where C is 200 meters.

To find the optimum values of L and D, we need to express one variable in terms of the other using the given constraints. Starting with the conditions L = 0, D = C/2, and the area being zero, we can see that as L increases, D decreases, and vice versa.

By setting D = C/2 when L = 0 and L = C/2 when D = 0, we establish the boundaries for our exploration. The midpoint of these boundaries, L = C/4 and D = C/4, represents the optimum solution.

The area A of the rectangle is given by A = L * D. Substituting L = C/4 and D = C/4 into the area formula, we get A = (C/4) * (C/4) = C^2/16. This yields the maximum area under the given constraints. The reason for this optimum lies in the balanced distribution of the perimeter, as equal values for L and D result in the most efficient use of the available length.