A rectangle has a perimeter of 24m. If its area is to be maximum, find its dimension.

To find the dimensions of a rectangle that has the maximum area with a given perimeter, we can start by expressing the perimeter in terms of the rectangle's dimensions. Let's call the length of the rectangle l and its width w. Then the perimeter P is given by: P = 2l + 2w We know that P = 24m, so we can substitute this into the above equation to get: 24 = 2l + 2w Simplifying this equation, we get: 12 = l + w Now we need to express the area of the rectangle in terms of l and w. The area A is given by: A = lw We want to find the values of l and w that will maximize A, subject to the constraint that 12 = l + w. One way to do this is to express l in terms of w using the equation 12 = l + w, and substitute this expression into the equation for A. We get: l = 12 - w A = lw = w(12 - w) = 12w - w^2 Now we have an expression for A in terms of just one variable, w. To find the maximum value of A, we can take the derivative of A with respect to w, set it equal to zero, and solve for w. We get: dA/dw = 12 - 2w = 0 w = 6 So the width of the rectangle that maximizes the area is 6m. To find the length, we can use the equation 12 = l + w, which gives us: l = 12 - w = 12 - 6 = 6 So the dimensions of the rectangle that has the maximum area with a perimeter of 24m are 6m by 6m. Therefore, the correct answer is 6, 6.