AS June 2022 Q4

EdexcelAS paperCurrent spec9 marksLinear Programming

Figure 3: graph for x from 0 to 16 and y from 0 to 14 showing three constraint lines through (0, 14) and (14, 0), through (0, 6) and (16, 14), and through (5, 0); the unshaded feasible region R is next to the origin, and an objective line runs from (0, 4) to (10, 0) — Figure 3

Figure 3 shows the constraints of a maximisation linear programming problem in \(x\) and \(y\), where \(x \geqslant 0\) and \(y \geqslant 0\). The unshaded area, including its boundaries, forms the feasible region, \(R\). An objective line has been drawn and labelled on the graph.

(a) List the constraints as simplified inequalities with integer coefficients. (3)

The optimal value of the objective function is 216

(b)

(i) Calculate the exact coordinates of the optimal vertex.

(ii) Hence derive the objective function. (5)

Given that \(x\) represents the number of small flower pots and \(y\) represents the number of large flower pots supplied to a customer,

Mark scheme (a)

Scheme	Marks	AO
\(x + y \leqslant 14\) \(2y - x \leqslant 12\) \(3x - y \leqslant 15\) \((x \geqslant 0,\ y \geqslant 0)\)	M1 A1 A1	3.3 1.1b 2.5
	(3)

Notes

M1: One correct non-trivial inequality in any form e.g. \(x - 2y + 12 \geqslant 0\). Condone strict inequality. Must be simplified to three terms only but coefficients do not need to be integers

A1: Two correct non-trivial inequalities in any form e.g. \(x - 2y + 12 \geqslant 0\). Condone strict inequalities. Must be simplified to three terms only but coefficients do not need to be integers

A1: All three non-trivial inequalities correct with three terms and integer coefficients

Mark scheme (b)

Scheme	Marks	AO
(i) Attempts to solve two equations to find optimal vertex	M1	3.4
\(\left(\dfrac{16}{3},\ \dfrac{26}{3}\right)\)	A1	1.1b
(ii) \(P = k(4x + 10y)\)	M1	3.1a
\(216 = k\left(4 \times \dfrac{16}{3} + 10 \times \dfrac{26}{3}\right)\)	ddM1	3.4
\((P =)\ 8x + 20y\)	A1	2.2a
	(5)

Notes

(b)(i) M1: Attempt to solve their \(x + y = 14\) and \(2y - x = 12\) (so their line with negative gradient and their line that passes through (0, 6)) simultaneously with at least one equation correct – the correct answer with no working implies this mark

A1: cao \(\left(\dfrac{16}{3},\ \dfrac{26}{3}\right)\) or \(\left(5\dfrac{1}{3},\ 8\dfrac{2}{3}\right)\) - must be exact (allow \(x = \ldots,\ y = \ldots\)) and clearly stated as the optimal vertex if more than one vertex of the FR found

(b)(ii) M1: Expression comprising of a constant (unknown) multiple/factor of \(2x + 5y\) e.g. \(k(4x + 10y)\) - M0 if assuming the objective is \(4x + 10y\) or if no \(k\) (or equivalent letter)

ddM1: Dependent on both previous M marks. Forming an equation with the expression \(k(4x + 10y)\) (or any multiple/factor of this), the 216 and their optimal vertex

A1: cao – accept \(8x + 20y\) or this expression equal to any letter but not for e.g. \(8x + 20y = 0\) or 216

Scheme	Marks	AO
6 small (flower pots) and 8 large (flower pots)	B1	3.2a
	(1)
	(9 marks)