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 a corral. If there is 200 feet of fencing available, what is the maximum possible area of the corral?
divide the fencing equally between lengths and widths.
Maximum area is this 100x50 = 5000 ft^2
see related problems below for the calculations that show this
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 a corral. If there is 760 feet of fencing available, what is the maximum possible area of the corral?
To find the maximum possible area of the corral, we need to maximize the length of the fence that can be used as the perimeter of the rectangle. Since the back of Dante's property is a creek and already acts as one side of the corral, we can allocate all of our available fencing to the other three sides.
Let's assume the length of the rectangle is L and the width is W. We can then write the equation for the perimeter:
2L + W = 200
To maximize the area, we need to solve for one variable in terms of the other. Let's solve the equation for W:
W = 200 - 2L
Now we can substitute this expression for W into the formula for calculating the area of a rectangle:
A = L * W
A = L * (200 - 2L)
To find the maximum area, we need to find the value of L that maximizes the equation A = L * (200 - 2L). We can do this by finding the vertex of the parabolic equation.
The vertex of a parabola in the form y = ax^2 + bx + c is given by (-b/2a, f(-b/2a)).
In our case, a = -2, b = 200, and c = 0. Plugging these values in, we get:
L_max = -200 / (2 * -2) = 50
Now we can find the width:
W_max = 200 - 2 * 50 = 100
Finally, we can calculate the maximum possible area:
A_max = L_max * W_max = 50 * 100 = 5000 square feet
Therefore, the maximum possible area of the corral is 5000 square feet.