The Pizza Connection calculates that the cost, in dollars, of producing \(x\) pizzas per day is given by

\begin{gather*}
C = 0.15x^2 + 0.75x + 180
\end{gather*}

The Pizza Connection charges $15 per pizza, so the revenue from selling \(x\) pizzas is

\begin{gather*}
R=15x
\end{gather*}

How many pizzas per day must the Pizza Connection sell in order to break even?

## Solution.

To break even means to make zero profit. Because

\begin{gather*}
\blert{\text{Profit} = \text{Revenue} - \text{Cost}}
\end{gather*}

the break-even points occur when revenue equals cost. In mathematical terms, we would like to find any values of \(x\) for which \(R = C\text{.}\) If we graph the revenue and cost functions on the same axes, these values correspond to points where the two graphs intersect. Use the

`WINDOW`settings
\begin{align*}
\text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 94\\
\text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 1400
\end{align*}

on your calculator to obtain the graph shown below. You can verify that the two intersection points are \((15, 225)\) and \((80, 1200)\text{.}\)

Thus, the Pizza Connection must sell either 15 or 80 pizzas in order to break even. On the graph we see that revenue is greater than cost for \(x\)-values between 15 and 80, so the Pizza Connection will make a profit if it sells between 15 and 80 pizzas.

We can also solve algebraically for the break-even points. The intersection points of the two graphs correspond to the solutions of the system of equations

\begin{align*}
y\amp =0.15x^2 + 0.75x + 180\\
y \amp= 15x
\end{align*}

We equate the two expressions for \(y\) and solve for \(x\text{:}\)

\begin{align*}
0.15x^2 + 0.75x + 180 \amp= 15x\amp\amp \blert{\text{Subtract }15x \text{ from both sides.}}\\
0.15x^2 - 14.25x + 180 \amp= 0\amp\amp \blert{\text{Use the quadratic formula.}}
\end{align*}

\begin{align*}
x \amp= \frac{14.25 \pm\sqrt{14.25^2 - 4(0.15)(180)}}{2(-0.05)}\amp\amp \blert{\text{Simplify.}}\\
\amp= \frac{14.25 \pm 9.75}{0.3}
\end{align*}

The solutions are 15 and 80, as we found from the graph.