18M.1.SL.TZ2.T_13

pestleMathematicsAASLPaper 118M· sl-5-8-testing-for-max-and-min-optimisation-points-of-inflexionsource ↗

A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by

C(x) = (x − 75)² + 100.

The cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.

Find the cost of producing 70 shirts.

[2]

Find the value of s.

[2]

Find the number of shirts produced when the cost of production is lowest.

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(70 − 75)² + 100 (M1)

Note: Award (M1) for substituting in x = 70.

125 (A1) (C2)

[2 marks]

(s − 75)² + 100 = 500 (M1)

Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.

(M1)

Note: Award (M1) for sketching correct graph(s).

(s =) 95 (A1) (C2)

[2 marks]

(M1)

Note: Award (M1) for an attempt at finding the minimum point using graph.

$\frac{{95 + 55}}{2}$ (M1)

Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.

(C'(x) =) 2x − 150 = 0 (M1)

Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.

75 (A1) (C2)

[2 marks]

Examiners’ report

[N/A]