Question

A(x)=− 4 1 ​ (x−25) 2 +625A, left parenthesis, x, right parenthesis, equals, minus, start fraction, 1, divided by, 4, end fraction, left parenthesis, x, minus, 25, right parenthesis, squared, plus, 625 The area, � ( � ) A(x)A, left parenthesis, x, right parenthesis, of a rectangular enclosure that can be made from a limited amount of fencing is shown above, where � xx is the length of one of the sides of the enclosure, measured in feet. What is the maximum area that can be enclosed in square feet?

Answer 1

Step-by-step explanation:

The given equation represents the area A(x) of a rectangular enclosure with one side length x, measured in feet. To find the maximum area that can be enclosed, we need to determine the value of x that maximizes A(x).

In this case, the equation for A(x) is a quadratic function in the form A(x) = -(1/4)(x - 25)^2 + 625. Since the quadratic term has a negative coefficient, the graph of this function will be a downward-opening parabola.

The maximum area will occur at the vertex of the parabola, which corresponds to the value of x that maximizes A(x). The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of the quadratic function.

In this case, a = -(1/4) and b = -25. Plugging these values into the formula, we get:

x = -(-25) / (2 * -(1/4))

x = 25 / (2/4)

x = 25 / (1/2)

x = 25 * (2/1)

x = 50

Therefore, the maximum area that can be enclosed in square feet is obtained when x = 50. We can substitute this value back into the equation for A(x) to find the maximum area:

A(50) = -(1/4)(50 - 25)^2 + 625

A(50) = -(1/4)(25)^2 + 625

A(50) = -(1/4)(625) + 625

A(50) = -156.25 + 625

A(50) = 468.75

Hence, the maximum area that can be enclosed is 468.75 square feet.