Combine meaningful functions into a quadratic decision

Try

A synthetic event model predicts n(p)=120−4p tickets sold at price p dollars per ticket, for 0≤p≤30. Revenue is price multiplied by ticket count, not the sum of those quantities. A fixed event cost is 180 dollars and each sold ticket adds 2 dollars of service cost. Decide whether maximizing revenue necessarily maximizes profit.

This is a continuous planning model, so fractional predicted sales may represent a mean forecast rather than literal fractional attendees. If prices must be whole dollars, that is a separate domain restriction. The linear demand relationship is assumed, not established by these calculations.

Predict

At p=10, predict sales, revenue, cost and profit before checking the table. Which quantities are measured in dollars and which in tickets? Explain why adding price directly to ticket count is meaningless here. Then predict whether reducing attendance can ever increase profit when it also reduces service cost.

Experiment

Price p, dollars/ticket	Tickets n(p)	Revenue R(p), dollars	Cost C(p), dollars	Profit P(p), dollars
0	120	0	420	−420
10	80	800	340	460
15	60	900	300	600
16	56	896	292	604
30	0	0	180	−180

Read both peak locations from the supplied graph, then verify their coordinates by rewriting the functions. The table supplies exact model values, not measured event outcomes. The shared vertical scale lets you compare dollars without comparing unrelated panel heights.

Understand

Revenue is the product R(p)=p·n(p)=120p−4p²=4p(30−p). The factored form reveals zero revenue at prices 0 and 30. Completing the square gives R(p)=900−4(p−15)², so its maximum on the stated domain is 900 dollars at p=15. The squared term is nonnegative, and the vertex input lies in the domain; both facts are needed for this conclusion.

Cost is a sum of meaningful components: C(p)=180+2n(p)=420−8p. The coefficient 2 carries dollars-per-ticket units; multiplying it by ticket count produces dollars compatible with the fixed cost. Profit subtracts functions with the same output units: P(p)=R(p)−C(p)=−4p²+128p−420=604−4(p−16)². Its maximum is 604 dollars at p=16. Maximizing revenue instead gives 600 dollars profit at p=15, so the objectives differ even in this simple model.

To require profit at least 500 dollars, solve 604−4(p−16)²≥500. This is equivalent to (p−16)²≤26, or 16−√26≤p≤16+√26. The interval is approximately [10.90,21.10], already inside [0,30]. If only integer-dollar prices are permitted, the qualifying prices are 11 through 21 inclusive. Round an endpoint only after determining the exact inequality; simply rounding a continuous optimum does not solve every discrete constraint problem.

The algebraic formula also exists for p>30, but then its sales prediction becomes negative. Those inputs are excluded from this contextual model. Likewise, negative profit means modeled revenue is below cost, not that a negative number of tickets was sold. A graph needs quantity labels and a domain before its sign or intercepts can be interpreted.

Apply

Build a similar model with changed demand and costs in practice. Write the component functions first, explain whether to add, subtract or multiply, and verify units before expanding. Report both continuous mathematical conclusions and any discrete purchasing or pricing decision. A supported coordinate description may replace a drawing, but must include the relevant vertices, intercepts, range and domain.

For a different quadratic context, a rectangle with side lengths x and 12−x has area x(12−x)=36−(x−6)². Proper rectangles require 0<x<12; the endpoint zero areas belong only to a model allowing degenerate rectangles. The same algebra has a different interpretation and different units.

Continue

Factored form exposes zeros, while vertex form exposes extrema; neither is universally the best representation. Combining functions requires meaningful quantities and a common input domain. These tasks develop F-IF.C.8a and F-BF.A.1 actions without claiming that a hypothetical demand curve has been validated or that this short unit is complete course preparation.