Answer:
(a) w = 75 ft and l = 46.875 ft
(b) A_max = 14062.5 ft²
Explanation:
To determine the optimal dimensions of the four equal pens created from the 750 ft of fencing, we will employ principles of optimization from calculus. We will first formulate an expression for the area in terms of one variable and then differentiate to find the dimensions that maximize the area. The parts of the problem involve determining the pen dimensions (a) and then computing the maximum overall area (b) using those dimensions.

Part (a) - Dimensions of Each Pen:

Let's denote:
- 'x' as the width of each pen,
- 'y' as the length of each pen.
Now, visualize the fenced area: It will have 3 internal fences dividing the 4 pens, plus the 2 external sides. Thus, for the width we will have 5 segments of width (3 inside + 2 outside). Similarly, there are 2 segments of length (top and bottom).
Therefore, the total fencing is:

From the above equation, we can express 'y' in terms of 'x':

Next, the area 'A' of all 4 pens combined is:

Substitute for 'y' using the relationship derived above:

To maximize the area, differentiate 'A(x)' with respect to x and set it to 0:

Setting 'dA/dx' equal to zero to find critical points:

Substitute 'x = 75' into the equation for 'y':

Therefore, the dimensions of the entire pen that will maximize the area are:
Width, 'x = 75 ft' and Length, 'y = 187.5 ft'.
To get the dimension of each pen seperately, simply divide the 'y' value by a factor of 4.
Thus, each pen has the following dimensions:
Width, 'w = 75 ft' and Length, 'l = 46.875 ft'.

Part (b) - Maximum Overall Area:

Using the determined dimensions, the maximum overall area 'A_max'
for all four pens combined is:

Therefore, the overall maximum area of the pen is 14062.5 ft².