Comparing Maximums between Quadratics

9.11.3 Comparing Maximums between Quadratics

9.11.3 • Comparing Maximums between Quadratics

Activity

A function $A$ , defined by $p (600 - 75 p)$ , describes the revenue collected from the sales of tickets for Performance A, a musical.

The graph represents a function $B$ that models the revenue collected from the sales of tickets for Performance B, a Shakespearean comedy.

A parabola on a coordinate grid representing function B. The x-axis represents the ticket price in dollars and the y-axis represents the revenue in dollars. The x-axis scale is 1 and extends from 0 to 8. The y-axis scale is 100 and extends from 0 to 1,400.

In both functions, $p$ represents the price of one ticket, and both revenues and prices are measured in dollars.

Without creating a graph of $A$ , determine which performance gives the greater maximum revenue when tickets are $p$ dollars each. Explain or show your reasoning.

Compare your answer:

Performance A.

The maximum revenue for Performance B is $900 based on the vertex (of a graph) of its graph. For Performance A, the expression $p (600 - 75 p)$ is in factored form, which tells us that the horizontal intercepts are $(0, 0)$ and $(8, 0)$ . The horizontal coordinate of the vertex is halfway between 0 and 8, which is 4. Substituting 4 for $p$ in $p (600 - 75 p)$ gives 1200, so the vertex is at $(4, 1200)$ . 1200 is greater than 900.
The maximum revenue for Performance B is $900 based on the vertex of its graph. The expression $p (600 - 75 p)$ can be rewritten in vertex form (of a quadratic expression): $- 75 (p - 4)^{2} + 1200$ , which tells us that the vertex of a graph of the function is at $(4, 1200)$ and that the maximum revenue is $1200.

Video: Comparing Maximums between Quadratics

Watch the following video to learn more about comparing maximums between quadratics.

Access multimedia content

Self Check

A function, defined by $p (600 - 60 p)$ , describes the revenue collected from the sales of tickets for a concert where $p$ is the price of one ticket. Find the maximum revenue, in dollars.

1500
10
5
600

Additional Resources

Finding Maximum Revenue

The revenue brought in by a newspaper is modeled by $R = p (- 2500 p + 159, 000)$ . Find the maximum revenue where $R$ is revenue and $p$ is price per newspaper subscription.

Step 1 - Find the horizontal intercepts or $x$ -intercepts (where $p = 0$ ).

First intercept -

This occurs at $p = 0$ or $(0, 0)$ .

Second intercept -

$\begin{matrix} - 2500 p + 159, 000 & = & 0 \\ - 2500 p & = & - 159, 000 \\ p & = & 63.6 \end{matrix}$

Step 2 - Find the $x$ -coordinate of the vertex halfway through the horizontal intercepts.

The vertex (of a graph) will occur halfway between $p = 0$ and $p = 63.6$ . This occurs when $p = 31.8$ .

Step 3 - Write in the function in standard form and substitute $p = 31.8$ .

$R = - 2500 p^{2} + 159, 000 p$

$R = - 2500 (31.8)^{2} + 159, 000 (31.8)$

The maximum revenue is at $2,528,100 with a $31.80 subscription.

A parabola on a coordinate grid. The vertex (31.80, 2,528.1) has been labeled on the parabola. The x-axis represents the price, p, in dollars and the y-axis represents the revenue in thousands of dollars. The x-axis scale is 1 and extends from 0 to 8. The y-axis scale is 500 and extends from 0 to 3,000.

Try it

Finding Maximum Revenue

Suppose that the price per unit in dollars of a cell phone production is modeled by $p = $ 45 - 0.0125 x$ , where $x$ is in thousands of phones produced, and the revenue represented by thousands of dollars is $R = x \cdot p$ . Find the production level that will maximize revenue and the maximum revenue.

Here is how to find the maximum revenue:

Step 1 - Find the horizontal intercepts.

$x = 0$ and $p = 45 - 0.0125 x$ $x = 3600$

Step 2 - Find the $x$ -coordinate of the vertex halfway through the horizontal intercepts.

This is the number of phones produced for the maximum revenue.

$x = 1800$

Step 3 - Find the equation for $R$ and substitute $x = 1800$ to find the maximum revenue.

$R = - 0.0125 x^{2} + 45 x$

$R = - 0.0125 (1800)^{2} + 45 (1800))$

$R = $ 40, 500$