Answer
Maximum area: $20, 833.33$sq ft
Work Step by Step
*Reference back to page 340 for a step-by-step solution to finding maximum and minimum values of quadratic functions.*
For this exercise, the first step is to identify what value the question seeks (maximum or minimum) and to write a function that expresses the variable in question in a quadratic function. Since we wish to maximize the AREA of the corral with length $x$ and width $y$, we can write the following:
$$Area = xy$$
We also know that the total length of fencing equals $1000$ft and that the inner fences run parallel to the corral borders. This means that there are 3 fences of length $x$ and 4 fences of length $y$ on the diagram. This allows us to create the following equation:
$$TotalFence = 1,000 = 3x + 4y$$.
Because we want to write a function in terms of one variable, we can use our second equation to solve for $y$ and plug that value into our first equation:
$$1,000 - 3x = 4y$$
$$\frac{1,000 - 3x}{4} = \frac{4y}{4}$$
$$250 - \frac{3}{4}x = y$$
Therefore:
$$Area = x(250 - \frac{3}{4}x)$$
$$Area = 250x - \frac{3}{4}x^{2}$$
Since we know that, in a quadratic function where $a<0$, its maximum value can be calculated when $x=\frac{-b}{2a}$, we can re-write our equation for AREA into a quadratic function with form $f(x) = ax^{2} + bx + c$:
$$f_{Area} = -\frac{3}{4}x^{2} + 250x + 0$$
We can now calculate the value of $x$ where the area is maximized:
$$x = \frac{-b}{2a} = \frac{-250}{2(- 3/4)} = \frac{500}{3}$$
Finally, we can substitute $x$ for $\frac{500}{3}$ to find the maximum AREA of the corral:
$$f_{maximum}(\frac{500}{3}) = -\frac{3}{4}(\frac{500}{3})^{2} + 250(\frac{500}{3})$$
$$f_{maximum}(\frac{500}{3}) = -\frac{3}{4}(\frac{250,000}{9}) + (\frac{125,000}{3})$$
$$f_{maximum}(\frac{500}{3}) = -\frac{62,500}{3} + \frac{125,000}{3}$$
$$f_{maximum}(\frac{500}{3}) = \frac{62,500}{3} \approx 20,833.33$$
