A farmer wants to enclose a rectangular field. He has a certain amount of fencing available, and he wants to maximize the area of the field he can enclose with it. This problem is a classic example of optimization in mathematics, where the goal is to find the best possible solution given certain constraints.
The farmer knows that the length of the field should be twice its width to maximize the area. However, he is unsure of the exact dimensions that will allow him to enclose the largest possible area within his budget. To solve this problem, the farmer needs to apply the principles of calculus and algebra.
Let’s denote the width of the field as ‘w’ and the length as ‘l’. Since the length should be twice the width, we can express the length as ‘2w’. The perimeter of the field, which is the total length of the fencing required, is given by the formula: P = 2l + 2w. Substituting the expression for ‘l’, we get: P = 2(2w) + 2w = 6w.
Now, the farmer wants to maximize the area of the field, which is given by the formula: A = lw. Substituting the expression for ‘l’, we get: A = 2w^2. Since the farmer has a limited amount of fencing, the perimeter constraint becomes: P = 6w = 6 (P/6) = P. This implies that the perimeter is equal to the amount of fencing available.
To find the maximum area, we need to differentiate the area formula with respect to ‘w’ and set it equal to zero. This will give us the critical points where the area is either maximized or minimized. Differentiating A = 2w^2 with respect to ‘w’, we get: dA/dw = 4w. Setting this equal to zero, we get: 4w = 0, which implies that w = 0. However, a width of zero is not a valid solution in this context.
To find the maximum area, we can use the second derivative test. The second derivative of the area formula is d^2A/dw^2 = 4. Since the second derivative is positive, the critical point we found (w = 0) corresponds to a minimum, not a maximum. This means that the maximum area occurs at a different width.
To find the width that maximizes the area, we can use the fact that the perimeter is equal to the amount of fencing available. Let’s assume the farmer has ‘F’ feet of fencing. Then, P = 6w = F. Solving for ‘w’, we get: w = F/6. Substituting this value of ‘w’ into the area formula, we get: A = 2(F/6)^2 = F^2/18.
Therefore, the maximum area the farmer can enclose is F^2/18 square feet. To find the corresponding length, we can use the expression for ‘l’: l = 2w = 2(F/6) = F/3. Thus, the farmer should enclose a rectangular field with a width of F/6 feet and a length of F/3 feet to maximize the area.
