O-Level Quadratic functions: worked solution

Question

Find the greatest number of pairs of shoes that can be made for a cost of $50 thousand.

The cost model from the earlier part of this question is C = 1.2(n − 6)² + 10.5, where C is the cost in thousands of dollars of making n hundred pairs of shoes. Put C = 50 (thousand dollars): 1.2(n − 6)² + 10.5 = 50 Subtract 10.5 from both sides: 1.2(n − 6)² = 39.5 Divide both sides by 1.2: (n − 6)² = 39.5 ÷ 1.2 = 32.9166... Take the square root of both sides, remembering both signs: n − 6 = ±√32.9166... = ±5.7373 n = 6 + 5.7373 = 11.7373 or n = 6 − 5.7373 = 0.2627 We want the greatest number of pairs, so take the larger root, n = 11.7373 (that is n ≈ 11.7 hundred). n is measured in hundreds, so the number of pairs = 11.7373 × 100 = 1173.73 pairs. A pair of shoes is a whole item, and 1174 pairs would cost more than 50 thousand dollars, so round down. Greatest number of pairs = 1173.