A Piecewise Pricing Model and an Invalid Linear Shortcut
A print shop charges $8 per shirt for orders up to 50 shirts and $6 per shirt for each shirt beyond 50, with a maximum order of 120 shirts. Fixed setup cost is $150, and blank shirts cost the shop $3.50 each. A student writes the profit function P(x)=4.5x−150 and concludes the shop maximizes profit by taking the largest possible order, 120 shirts. The student's table shows P(120)=$390. The shop owner says that at 120 shirts the actual profit is lower than the student's prediction but still positive. You ask the class to determine where the linear model stops representing the pricing rule.
1. Write the correct piecewise profit function and compute the actual profit at 120 shirts. How large is the student's error?
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For 0≤x≤50, revenue is 8x, so P(x)=8x−(3.5x+150)=4.5x−150. For 50<x≤120, revenue is 8·50+6(x−50)=6x+100, so P(x)=6x+100−(3.5x+150)=2.5x−50. At x=50 both pieces give $75, so the function is continuous. At x=120, P(120)=2.5·120−50=250, not 390. The student overestimates profit by $140. The student's single linear rule ignores the price drop after 50 shirts and therefore gives a revenue slope of 8 rather than 6 on the second interval.
2. The student argues that the linear model is acceptable because it still predicts that profit increases with order size. Evaluate that reasoning.
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The conclusion happens to remain true: on 50<x≤120 the correct slope is 2.5, which is positive, so the maximum still occurs at 120 shirts. But a correct qualitative direction does not validate the model quantitatively. The student's predicted maximum is $390 instead of $250, a 56% overstatement, and any threshold decision can change. For example, the order size needed for $200 profit is different under the two models. A model must be checked against the actual piecewise structure and domain, not merely against the sign of its slope.
3. The owner wants a profit of at least $200. Using the correct model, what order sizes qualify, and how would the student's model mislead the owner?
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For x≤50 the largest possible profit is 4.5·50−150=$75, which is below $200, so the qualifying order must exceed 50 shirts. On that interval, 2.5x−50≥200 gives 2.5x≥250, so x≥100. With the 120-shirt maximum, qualifying orders are 100 through 120 shirts. The student's model gives 4.5x−150≥200, so x≥77.8, or 78 shirts. The owner following the student's model could accept an order of 78 shirts and earn only 2.5·78−50=$145, well short of the target.