The back of Tom's property is a creek. Tom would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create a corral. If there is 100 feet of fencing available, what is the maximum possible area of the corral?
? square feet ?
1250
length of the 1 side parallel to the creek --- y
length of each of the 2 other sides ---- x
2x + y = 100
y = 100-2x
area = xy
= x(100-2x)
= -2x^2 + 100x
this is modelled by a downwards opening parabola.
Find the vertex and you got the answer.
Well, if Tom's property is anything like mine, then the creek probably has some interesting creatures in it. Maybe he can turn the corral into a petting zoo and charge admission. That way, he can use the money to buy even more fencing to expand the corral. It's a win-win, and the area of the corral will grow along with his profits!
To find the maximum possible area of the corral using the given 100 feet of fencing, we need to determine the dimensions of the rectangular area that would give us the largest possible area.
Let's assume that the creek side of the corral is the length of the rectangle, denoted by "l". The other two sides will be the width, denoted by "w".
Since we have 100 feet of fencing available, the total length of the three sides (excluding the creek) will be 100 feet. This means that the total length of the fencing on the three sides will be l + 2w.
To maximize the area, we want to find the dimensions that would give us the largest value for l and w.
Since the creek is already fixed as one side of the rectangle, its length (l) is already determined and does not affect the total length of the fencing (l + 2w). Therefore, we want to allocate the remaining 100 feet of fencing equally between the two widths.
Let's say we allocate "x" feet to each width (w). This means that the remaining fence length for the length (l) side would be (100 - 2x) feet.
The area (A) of the rectangle can be calculated by multiplying the length (l) and width (w): A = l * w.
Substituting the expressions for l and w using the allocated fence lengths, we get:
A = (100 - 2x) * x.
To find the maximum possible area, we need to find the value of x that maximizes this equation. We can do this by finding the critical points of the expression and determining whether they correspond to a maximum.
Taking the derivative of the expression with respect to x, we get:
A' = 100 - 4x.
To find the critical points, we set the derivative equal to zero:
100 - 4x = 0.
Solving for x, we get:
4x = 100.
x = 25.
To verify that this is a maximum and not a minimum or an inflection point, we would need to take the second derivative. However, since this is a quadratic equation, we can observe that the coefficient of x^2 is negative, which confirms that x = 25 corresponds to a maximum.
So, the width (w) of the corral will be 25 feet, and the length (l) will be 100 - 2x = 100 - 2*25 = 50 feet.
Therefore, the maximum possible area of the corral is A = l * w = 50 * 25 = 1250 square feet.