Question

300 ft of fencing will be used to enclose a rectangular corral split into 2 pens of equal size. find the dimensions of the corral that will enclose the greatest possible area

Answer 1

25 feet by 16 2/3 feet separated into two smaller pens of 12 1/2 feet by 16 2/3 feet.

I can see two potential ways of interpreting this question:
1. 300 feet of fencing will be used to enclose an area and the area will be subdivided in half using some other method. This interpretation is rather trivial since you would just need a square with each side equal to 300/4 = 75 feet.

2. 300 feet of fencing will be used to both enclose the area and provide the subdividing. This is a harder problem and will be the one I solve.

First, let's create two variables, W and L. I'll need three lengths of fencing W feet long in order to handle both ends of the rectangle plus the division in the middle. I'll also need 2 lengths of fencing L units long to handle the top and the bottom. So
3W + 2L = 100
and the total area will be
A = WL

Now let's express L in terms of W
3W + 2L = 100
2L = 100 - 3W
L = (100 - 3W)/2
L = 50 - (3/2)W

And express area in terms of W
A = WL
A = W(50 - (3/2)W)
A = 50W - (3/2)W^2
A = -(3/2)W^2 + 50W

Using the quadratic formula, you get the roots 0 and 33 1/3
Since a quadratic equation makes a parabola and since a parabola is symmetrical, the maximum should happen midway between the two roots, which would be 16 2/3 or 50/3. Let's check if that's correct by using (e + 50/3) as the width and see what we get.
A = -(3/2)W^2 + 50W
A = -(3/2)(e + 50/3)^2 + 50(e + 50/3)
A = -(3/2)(e^2 + (100/3)e + 2500/9) + 50e + 2500/3
A = -(3/2)e^2 -50e -1250/3 + 50e + 2500/3
A = -(3/2)e^2 +1250/3

Notice the only term that has e. It's -(3/2)e^2. If e has ANY value other than 0, it will cause that term to be negative and reduce the area of the enclosure. That means that our value of 50/3 for the width is the correct value. So let's get the length now.

L = 50 - (3/2)W
L = 50 - (3/2)(50/3)
L = 50 - 150/6
L = 50 - 25
L = 25

So the enclosure will be 25 feet by 16 2/3 feet separated into two smaller pens of 12 1/2 feet by 16 2/3 feet.

Answer 2

There are two possible interpretations of this question. They are:
1. Largest possible area to enclose with 300 ft of fencing and that area with be divided in two via some other means. The answer to this version is rather trivial in that the rectangle needs to be a square and the length of each side will be 300/4 = 75 feet.
2. Largest possible area to enclose and the division between the two areas will also be done with the same fencing. This problem isn't as trivial and will be the one that I answer.
First, I will consider the number of lengths of fencing. I see 3 for the width and 2 for the length. For the width, 2 will be the sides and the 3rd will be the divider. I'll use the values W for the width and L for the length. So
3W + 2L = 100
A = WL
Let's express L in terms of W
3W + 2L = 100
2L = 100 - 3W
L = (100 - 3W)/2
Now express Area in terms of W
A = WL
A = W(100 - 3W)/2
Distribute W and divide by 2
A = (100W - 3W^2)/2
A = 50W - 3/2W^2
A = -3/2W^2 + 50W
Since this is claimed to be a highschool problem, I'll avoid the use of calculus and simply note that the equation for area is a simple quadratic equation. I'll use the quadratic formula to get its two roots of 0 and 33 1/3. Since all quadratic equations describe a parabola and since a parabola is bilaterally symmetrical, let's pick a possible solution between it's roots. So the average of 0 and 33 1/3 is 16 2/3 or 50/3. Let's see if that value is actually the best possible. I'll use the value (50/3 + e) for W and see what I get when I multiply it out.
A = -3/2W^2 + 50W
A = -3/2(50/3 + e)^2 + 50(50/3 + e)
A = -3/2(e^2 + (100/3)e + 2500/9) + 2500/3 + 50e
A = (-3/2e^2 - 50e - 1250/9) + 2500/3 + 50e