## College Algebra (6th Edition)

Playground's dimensions: $100$ ft $\times 150$ ft Maximum area: $15,000$ sq feet
*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 rectangular playground ($length \times width$) whose total fencing must equal 600 feet after dividing the playground into 2. Let us assume that one pair of the playground's sides measures $x$ and the other measures $y$. Since we don't know which variable represents length and which represents width, we can arbitrarily assign the new fence between either pair. That way, we can say that: $$TotalFence = 600 = 3x + 2y$$ Since we need to write a function in terms of just one variable, we can solve for y in the following manner: $$600=3x+2y$$ $$600 - 3x = 2y$$ $$\frac{600 - 3x}{2} = \frac{2y}{2}$$ $$300 - \frac{3}{2}x = y$$ which means that one side of the playground should measure $x$ and the perpendicular side should measure ($300 - \frac{3}{2}x$). The area of the playground (which is the real value that the question is seeking), can now be represented as: $$Area = (x)(300-\frac{3}{2}x) = 300x - \frac{3}{2}x^{2}$$ 2) We can re-write the equation we made for area into a quadratic function with form $f(x)=ax2+bx+c$: $$f_{Area}(x)=−\frac{3}{2}x^{2} + 300x + 0$$ 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: $$x_{maximum}=\frac{-b}{2a}=\frac{-300}{2(-3/2)} = \frac{-300}{-3} = 100$$ 4) With the value of $x$ at hand, we can now determine the maximum area for the function we've established: $$f_{maximum}(x) = f(100)= 300(100) - \frac{3}{2}(100)^{2}$$ $$f(100)= 30,000 - \frac{3}{2}(10,000)$$ $$f(100) = 30,000 - 15,000 = 15,000$$ 5) Now that we have the value of the maximum area, the only value missing is the remaining pair of the playground's sides which we previously denominated as $y$, and which we can calculate using the formula we derived in step 1: $$y = 300 - \frac{3}{2}x_{maximum}$$ $$y = 300 - \frac{3}{2}(100) = 300 - 150 = 150$$