Linear Demand Extrapolation in Cafeteria Pricing
A high-school algebra team is analyzing cafeteria lunch sales after several price changes. The counts are: at $2.50, 400 meals sold; at $3.00, 340; at $3.50, 300; at $4.00, 270. A student uses the first two points to fit the linear demand model q = 700 − 120p and predicts 220 meals at $4.00. The actual count is 270. The student concludes that the price increase 'cost 180 meals' and recommends returning to $2.50 to maximize revenue. The teacher wants students to critique the model, compute residuals, and decide whether a linear, quadratic, or piecewise model is defensible from four data points. The class has graphing calculators and 45 minutes. The cafeteria manager cares about revenue, not meal count alone.
1. Compute the linear model's prediction at $4.00, the residual, and the revenue implied by the linear model at $2.50, $3.00, $3.50, and $4.00. What pattern shows the model is inadequate?
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Linear model: q = 700 − 120p. At p = 4.00, q = 700 − 480 = 220. Residual = actual − predicted = 270 − 220 = 50 meals; the model underpredicts by 50. Revenue R = pq at the four observed prices: 2.50 × 400 = 1000; 3.00 × 340 = 1020; 3.50 × 300 = 1050; 4.00 × 270 = 1080. The linear model's revenue at 3.50 would be 3.50 × 280 = 980, and at 4.00 it would be 4.00 × 220 = 880. Thus the linear model predicts revenue falling as price rises from 3.50 to 4.00, while the observed revenue is rising. The residual at 4.00 and the reversal in revenue pattern show the linear demand model is not adequate over this range.
2. Fit a quadratic demand model to the points at $2.50, $3.00, and $3.50, and use it to predict at $4.00. Compare with the actual count.
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Let x = p − 2.50. Use (0, 400), (0.5, 340), (1.0, 300). With q = ax² + bx + c, c = 400. At x = 0.5: 0.25a + 0.5b = −60, so a + 2b = −240. At x = 1: a + b = −100. Subtracting gives b = −140, then a = 40. So q = 40x² − 140x + 400. At p = 4.00, x = 1.5: q = 40(2.25) − 140(1.5) + 400 = 90 − 210 + 400 = 280. The actual is 270, so the quadratic residual is −10 (overpredicts by 10), much smaller in magnitude than the linear residual of 50. This does not prove the quadratic is true; it shows that a curved demand model fits the observed curvature better and that the linear extrapolation was unreliable.
3. What instructional point about interpolation versus extrapolation and about using revenue rather than demand should the teacher make? Give a specific recommendation.
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Interpolation within the observed range is safer than extrapolation beyond it. With four points, infinitely many models can fit exactly or nearly; the teacher should ask students to compare residuals and to avoid treating any one model as reality. Revenue is R(p) = p·q(p), not q(p) alone: at $4.00 the observed revenue is $1080, higher than $1000 at $2.50, so the recommendation to return to $2.50 does not follow. A defensible recommendation is to test prices near $4.00 with a controlled short experiment, because the data suggest revenue may still be increasing, but the model cannot tell how far. Students should state the assumption that the price range remains within the observed conditions.