What about regions built along a natural boundary? For example the maximum for both a rectangular region and a triangular region built along a natural boundary with 100 yards of fencing is 1250 sq. yds. But the rectangle is not the maximum area four-sided figure that can be built. What is the maximum-area four-sided figure?

Answer 1

For a four-sided figure, we should consider a trapezoid. Here's how we can do it:

Let's say the trapezoid is built alongside the natural boundary with its two bases parallel to the boundary. Let the lengths of the bases be b1 and b2, and let the height be h. To maximize the area of the trapezoid, the sum of the two non-parallel sides should be equal to the remaining fencing, which is 100 yards in this case.

Let x be the difference between the two bases, so that b2 = b1 + x. The area A of the trapezoid can be expressed as:

A = 0.5 * (b1 + b2) * h.

We can use the Pythagorean theorem to relate h and x to the length of the non-parallel sides of the trapezoid:

h^2 + (x/2)^2 = (50 - b1)^2.

Now, we substitute b2 = b1 + x in the area formula:

A = 0.5 * (b1 + b1 + x) * h = (b1 + 0.5x) * h.

We can solve for h from the Pythagorean theorem equation and substitute it into the area formula:

h = sqrt((50 - b1)^2 - (x/2)^2).

So, A = (b1 + 0.5x) * sqrt((50 - b1)^2 - (x/2)^2).

Now, to maximize the area A, we can take the derivative of A with respect to b1 and set the result equal to 0, and then solve for b1. We would also need to determine an expression for x in terms of b1. This would involve solving a system of equations and using calculus for optimization.

However, this approach is quite complex and, given the nature of the problem, is likely to require numerical methods for a solution. The maximum area for the trapezoid (a four-sided figure) built along the natural boundary can then be determined using those values of b1 and x. This value is expected to be greater than 1250 sq. yds, which is the maximum area for the rectangular and triangular regions mentioned.