Problem 66 - Algebra and Trigonometry

Question

You have 200 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

Step-by-Step Solution

Answer

The dimensions that will yield the largest enclosed area are a length of 100 feet and a width of 50 feet. The largest area that can be enclosed with the 200 feet of fencing is 5000 square feet.

Step 1: Create the function for Area

The first step is to model the problem mathematically, specifically to define the area of the plot. The area 'A' of a rectangle is the product of its length 'L' and its width 'W'. In our scenario, based on the length of the fencing, the Length 'L' can be expressed as \( L = 200 - 2W \). So the area can be written as: \( A = W \times (200 - 2W) \) which simplifies to \( A = 200W - 2W^2 \).

Step 2: Find the derivative of the Area

To find the maximum area, we need to find the critical points of 'A' by taking its derivative and setting it equal to 0. The derivative of A is \( A' = 200 - 4W \). Set this equal to zero and solve for 'W' to find the critical points: \( 200 - 4W = 0 \) hence \( W = 50 \).

Step 3: Verify the maximum

To verify that 'W' we found does indeed give a maximum area, we can substitute '50' into the second derivative of 'A', which is \( A'' = -4 \). Since \( A'' < 0 \), it confirms that this is indeed a maximum point.

Step 4: Calculate the Maximized Area

Now we substitute 'W' into our equation for 'A' to find the maximum enclosed area: \( A = 200(50) - 2(50)^2 = 5000 square feet \). So the largest enclosed area is 5000 square feet.

Step 5: Find the lengths

The length of the fence perpendicular to the river is given as 'W = 50 feet' and the other length 'L' is \( L = 200 - 2W = 200 - 2(50) = 100 feet \). So, the required dimensions are 100 feet by 50 feet.