Answer
$x_{maximum} = 12.5$ yards; $f(12.5) = 156.25$ 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 ($length \times width$) whose perimeter must equal 50 yards [($2\times length) + (2 \times width$)]. Let $x$ represent width and $y$ represent length for just a moment. We get the following:
$$Perimeter = 50 = 2x + 2y$$
Since we need to write a function in terms of just one variable, we can solve for $y$ in the following manner:
$$50 = 2x + 2y$$
$$50 - 2x = 2y$$
$$\frac{50 - 2x}{2} = \frac{2y}{2}$$
$$25x - x = y$$
which gives us the following:
$$Perimeter = 2x + 2(25x -x)$$
$$Area = (x)(25 - x) = 25x - x^{2}$$
2) We can re-write the equation we made for area into a quadratic function with form $f(x) = ax^{2} + bx + c$:
$$f_{Area}(x) = -x^{2} + 25x + 0$$
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:
$$x_{maximum} = \frac{-b}{2a} = \frac{-25}{2(-1)} = \frac{-25}{-2}$$
$$x_{maximum} = \frac{25}{2} = 12.5$$
4) With the value of $x$ at hand, we can now determine the maximum area for the function we've established:
$$f_{maximum}(x) = f(12.5) = 25(12.5) - (12.5^{2})$$
$$f_{maximum}(12.5) = 312.5 - (156.25) = 156.25$$
