Answer
$length = 20$ yards; $width = 20$ yards; $Area_{maximum} = 400$ sq yards
Work Step by Step
*Refer back to page 340 for a complete reference on calculating maximum and minimum values of quadratic functions*
1) The first step is to identify what value the question is asking for and to write it in a function form. They're asking for the maximum area of a rectangle (length×width) whose perimeter must equal 80 yards [(2×length)+(2×width)]. Let x represent width and y represent length for just a moment. We get the following:
$$Perimeter=80=2x+2y$$
Since we need to write a function in terms of just one variable, we can solve for y in the following manner:
$$80=2x+2y$$
$$80−2x=2y$$
$$\frac{80 - 2x}{2}=\frac{2y}{2}$$
$$40−x=y$$
which gives us the following:
$$Perimeter=2x+2(40−x)$$
$$Area=(x)(40−x)=40x−x^{2}$$
2) We can re-write the equation we made for area into a quadratic function with form $f(x)=ax2+bx+c$:
$$f_{Area}(x) = −x2 + 40x + 0$$
3) Since we know that, for quadratic equations with $a<0$, we can find the maximum value at $x=\frac{-b}{2a}$, we can determine the following:
$$x_{maximum}=\frac{-b}{2a} = \frac{-40}{2(-1)} = \frac{-40}{-2} = 20$$
4) With the value of $x$ at hand, we can now determine the maximum area for the function we've established:
$$f_{maximum}(x)=f(20)=40(20)−(20^{2})$$
$$f_{maximum}(20)=800 - 400 = 400$$
