Mathematics 2001 JAMB Past Questions - Question 37: Find the dimensions of the rectangle of greatest area which has a fixed perimeter p.

Correct Answer

A

Explanation

To find the dimensions of the rectangle with the greatest area, given a fixed perimeter p, we can use the concept of optimization.Let's denote the length of the rectangle as L and the width as W. The perimeter of the rectangle is given by the formula:p = 2L + 2WWe can rearrange this equation to solve for one of the variables in terms of the other:L = (p - 2W) / 2Now, the area of the rectangle is given by the formula:A = L * WSubstituting the expression for L, we have:A = [(p - 2W) / 2] * WSimplifying further:A = (pW - 2W²) / 2To find the dimensions of the rectangle that maximize the area, we need to find the critical points of the area function. We can do this by taking the derivative of A with respect to W and setting it equal to zero:dA/dW = (p - 4W) / 2 = 0p - 4W = 0W = p / 4Substituting this value of W back into the equation for L, we have:L = (p - 2(p/4)) / 2L = (p - p/2) / 2L = p / 4Therefore, the dimensions of the rectangle with the greatest area, given a fixed perimeter p, are L = p / 4 and W = p / 4.