Nora is creating a dog run for her dog, Mr. Darcey. She has enough fencing material for a fence with a perimeter of 120 feet. This situation is modeled using the equation A=−w^2+60w
where A
is the area and w is the width. Using this equation, identify the maximum range or area the fence can have.(1 point)
square feet
The equation given, A = -w^2 + 60w, represents a quadratic equation in the form of a parabola. In this equation, A stands for the area of the dog run in square feet, and w stands for the width in feet. We can find the maximum area by locating the vertex of this parabola. Since it's given in the vertex form A = a(w - h)^2 + k where (h, k) is the vertex of the parabola, we can re-write the equation (although unnecessary).
The standard form of a quadratic equation is ax^2 + bx + c. When comparing this to the equation we have, we can see that the coefficient of w^2 is -1 (a = -1), so the parabola opens downwards, which means it will have a maximum value at its vertex.
To find the h-coordinate of the vertex (which in this case would be the width that gives us the maximum area), we use the formula h = -b/(2a). However, since our equation is not in standard form, we can find the value that maximizes the area by taking the derivative of A with respect to w and set it equal to zero or by taking half of the coefficient of the w term in our equation (since this is a simpler quadratic parabola). In -w^2 + 60w, b = 60, and a = -1, so:
h = -b/(2a)
h = -60/(2 * -1)
h = -60 / -2
h = 30
This tells us that the width w of the rectangle that maximizes the area is 30 feet. To find the maximum area, we substitute w back into the equation:
A = -(30)^2 + 60(30)
A = -900 + 1800
A = 900 square feet
So, the maximum area Nora can fence for the dog run for her dog, Mr. Darcey, is 900 square feet.