Question

The back of Dante's property is a creek. Dante would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create acorral. If there is 760 feet of fencing available, what is the maximum possible area of the corral?

Answer 1

Let's draw a diagram of the situation:

According to the information of the problem, we have 760 feet of fencing available, which means

`2a+b=760.`

The area of this enclosure would be

`A=a\cdot b\text{.}`

Let's rewrite the equation of the perimeter we got earlier to make one of the variables depend on the other:

`b=760-2a\text{.}`

Now let's use this value of b in the formula for the area:

`A=a\cdot(760-2a)=760a-2a^2\text{.}`

We can consider this as a function of a:

`A(a)=a(760-2a)=-2a^2+760a\text{.}`

From

`-2a^2+760a,`

we can tell that the function is a parabola.

From

`A(a)=a(760-2a),`

we can tell that A(a)=0 when a=0 and when a=380. Since the leading term of the parabola is negative (-2) this means that we can think of the parabola starting at (0,0) going up, reaching a certain maximum and then going down to reach (0,380):

Since a parabola is simmetrical, we can conclude that it will reach its highest point at the midpoint between 0 and 380. In other words, it will reach its maximum when a=190.

And when a=190:

`A(190)=-2(190)^2+760(190)=-2(36,100)+144,400=-72,200+144,400=72,200.`

Thus, the maximum possible are of the enclosement is 72,200 square feet.