Question

3 a homeowner is planning a garden. one side of the garden is against a house, but the other sides need a fence. which graph can you use to determine the largest area the homeowner can have for the garden using 50 feet of fencing? a. b. c. d.

Explanation:

Step1: Let the length perpendicular to the house be $x$ and the length parallel to the house be $y$. The perimeter of the fenced - in area (using 50 feet of fencing) is $2x + y=50$, so $y = 50 - 2x$.

The area $A$ of the rectangle is $A=xy=x(50 - 2x)=-2x^{2}+50x$.

Step2: The function $A(x)=-2x^{2}+50x$ is a quadratic function in the form $y = ax^{2}+bx + c$ with $a=-2$, $b = 50$, and $c = 0$. The vertex of a quadratic function $y = ax^{2}+bx + c$ has its $x$ - coordinate at $x=-\frac{b}{2a}$.

For $A(x)=-2x^{2}+50x$, $x=-\frac{50}{2\times(-2)}=\frac{50}{4}=12.5$.

Step3: Substitute $x = 12.5$ into the area function $A(x)$: $A(12.5)=-2\times(12.5)^{2}+50\times12.5=-2\times156.25 + 625=-312.5+625 = 312.5$.

The maximum of the quadratic function $A(x)=-2x^{2}+50x$ occurs at $x = 12.5$ and $A = 312.5$. We need to find the graph with a maximum around $x = 12.5$ and $y=312.5$.

Answer:

b.