Marquise has $200$ meters of fencing to build a rectangular garden. The garden's area (in square meters) as a function of the garden's width (in meters) is modeled by: $A(x)=-x^2+100x$ What is the maximum area possible?

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Q. Marquise has

200

meters of fencing to build a rectangular garden. The garden's area (in square meters) as a function of the garden's width (in meters) is modeled by:

A(x)=-x^2+100x

What is the maximum area possible?

Perimeter Calculation: Marquise has $200$ meters of fencing to build a rectangular garden. The perimeter of a rectangle is given by $P = 2l + 2w$ , where $l$ is the length and $w$ is the width. Since Marquise has $200$ meters of fencing, we have $200 = 2l + 2w$ , or $100 = l + w$ . This means that the length can be expressed in terms of the width as $l = 100 - w$ .
Area Calculation: The area of a rectangle is given by $A = l \times w$ . Substituting $l = 100 - w$ into the area formula, we get $A(w) = (100 - w) \times w = 100w - w^2$ , which matches the given function $A(x) = -x^2 + 100x$ . This confirms that the function correctly represents the area of the garden in terms of its width.
Finding Maximum Area: To find the maximum area possible, we need to find the vertex of the parabola represented by the function $A(x) = -x^2 + 100x$ . Since the coefficient of $x^2$ is negative, the parabola opens downwards, and the vertex will give us the maximum area.
Calculating Vertex: The $x$ -coordinate of the vertex of a parabola in the form of $f(x) = ax^2 + bx + c$ is given by $-\frac{b}{2a}$ . In our case, $a = -1$ and $b = 100$ . Therefore, the $x$ -coordinate of the vertex is $-\frac{100}{2*(-1)} = 50$ .
Substitute for Maximum Area: Substitute $x = 50$ into the area function to find the maximum area. $A(50) = -50^2 + 100 \times 50 = -2500 + 5000 = 2500$ square meters.
Final Maximum Area: The maximum area possible for the garden with $200$ meters of fencing is $2500$ square meters.