Linear Cost Model Pushed Beyond Its Data
A print shop quotes a customer $290 for an order of 40 shirts and $650 for an order of 100 shirts of the same design, with one fixed setup charge and a constant per-shirt price. A student in a tutoring session models the total cost in dollars, C(n), for n shirts as C(n) = 6n + 50 and submits the model as complete. The customer then asks what an order of 2,000 shirts would cost so she can compare it with an online wholesaler, and the student substitutes n = 2,000 and reports $12,050. The tutor knows the shop gives an 8% discount on the per-shirt price for any order above 500 shirts, a fact the student never asked about. The tutor must decide what to correct: the algebra, the meaning of the parameters, or the use of the model outside the range of the two quotes.
1. Verify the student's model from the two quotes, then state what the slope and the vertical intercept mean in this context and whether each is plausible.
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Slope = (650 - 290)/(100 - 40) = 360/60 = 6 dollars per shirt. Substituting: 290 = 6(40) + b, so b = 50.
The model reproduces both quotes exactly, so it is correct as a summary of the two data points.
- Slope: each additional shirt adds $6 to the total, i.e., the marginal per-shirt price.
- Intercept: C(0) = $50, the fixed setup charge that is paid regardless of the number of shirts.
Both are plausible for a screen-printing shop: a one-time setup fee plus a per-unit charge is a standard cost structure, and a positive intercept is consistent with that structure. Note that a nonzero C(0) is a limiting value, not a real order - the shop would not charge $50 for zero shirts unless an order is actually placed. Also note that two points determine a line but do not prove the shop's pricing is linear; a third quote would test the assumption.
2. The student reports $12,050 for 2,000 shirts. Evaluate the reasoning behind that number and describe what a defensible response to the customer would be.
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The arithmetic is correct: 6(2,000) + 50 = 12,050. The error is in the modeling, not the computation. The student extended a line validated only on the interval [40, 100] to n = 2,000, implicitly assuming the per-shirt price never changes. The tutor knows it does change: above 500 shirts the per-shirt price drops 8%, to 6(0.92) = $5.52.
Using that rate with the same setup fee gives 50 + 5.52(2,000) = $11,090, a difference of $960 from the student's figure. Even this is an estimate, because the discount could apply only to the shirts above 500, or the shop could change the setup fee for large runs.
A defensible response to the customer: present $12,050 as an upper bound based on the quoted small-order pricing, state clearly that pricing above 500 shirts has not been confirmed, and get the shop's actual quote before the customer uses the number in a comparison. The tutor should also have the student record the domain on which the model was built.
3. State a reasonable domain for C(n) for the two original quotes and explain how the domain affects any claim that C is linear.
Read the worked answer
Domain: n must be a whole number of shirts, since fractional shirts cannot be ordered, and the model was built from orders of 40 and 100. A defensible domain is the integers 40 ≤ n ≤ 100 (or 1 ≤ n ≤ 500 if the shop confirms the same pricing over that whole band). The range is then the corresponding discrete set of dollar amounts, not a continuous interval.
The domain matters for the linearity claim in three ways:
- C is a discrete function of a whole-number input; drawing a continuous line through it is a convenience, not a claim that 40.5 shirts are sold.
- Linear behavior is an assumption supported by two points, not a verified fact; it holds on the quoted band only until another price break appears.
- Above 500 shirts the structure changes, so the single linear rule becomes a piecewise rule (discounted slope). The student's sentence 'the cost is 6n + 50' is therefore true only on a restricted interval, and the tutor should require that interval to be stated with the model.