D1 June 2011 Q3

EdexcelOld spec8 marksLinear Programming

Figure 2: graph with lines 6x + 5y = 60, 2x + 3y = 12, 3x = 2y and x = 2y, rejected sides hatched, feasible region R between them — Figure 2

Figure 2 shows the constraints of a linear programming problem in \(x\) and \(y\), where \(R\) is the feasible region.

(a) Write down the inequalities that form region \(R\). (2)

The objective is to maximise \(3x + y\).

(b) Find the optimal values of \(x\) and \(y\). You must make your method clear. (4)

Given that integer values of \(x\) and \(y\) are now required,

(d) write down the optimal values of \(x\) and \(y\). (1)

Mark scheme (a)

Scheme	Marks
\(6x + 5y \leqslant 60\) \(2x + 3y \geqslant 12\) \(3x \geqslant 2y\) \(x \leqslant 2y\)	B2,1,0
	(2)

Notes

(a)1B1 Any two inequalities correct, accept < and > here (but not = of course).

2B1 All four correct. Must be \(\leqslant\) and \(\geqslant\) here

Mark scheme (b)

Scheme	Marks
Drawing objective line{ (0,3) (1,0)} Testing at least 2 points	M1 A1
Calculating optimal point Testing at least 3 points	DM1
\(\left(7\dfrac{1}{17}, 3\dfrac{9}{17}\right) = \left(\dfrac{120}{17}, \dfrac{60}{17}\right) \approx (7.06, 3.53)\)	A1 awrt
	(4)

Notes

\[\begin{aligned}&\left(3\tfrac{3}{7}, 1\tfrac{5}{7}\right) = \left(\tfrac{24}{7}, \tfrac{12}{7}\right) \approx (3.43, 1.71) \rightarrow 12\\ &\left(1\tfrac{11}{13}, 2\tfrac{10}{13}\right) = \left(\tfrac{24}{13}, \tfrac{36}{13}\right) \approx (1.85, 2.77) \rightarrow 8.307692\ \left(8\tfrac{4}{13} = \tfrac{108}{13}\right)\\ &\left(4\tfrac{4}{9}, 6\tfrac{2}{3}\right) = \left(\tfrac{40}{9}, \tfrac{20}{3}\right) \approx (4.44, 6.67) \rightarrow 20\\ &\left(7\tfrac{1}{17}, 3\tfrac{9}{17}\right) = \left(\tfrac{120}{17}, \tfrac{60}{17}\right) \approx (7.06, 3.53) \rightarrow 24.705882\ \left(24\tfrac{12}{17} = \tfrac{420}{17}\right)\end{aligned}\]

(b)1M1 Drawing objective line or its reciprocal OR testing two vertices in the feasible region (see list above) points correct to 1 dp.

1A1 Correct objective line OR two points correctly tested (1 dp ok)

2DM1 Calculating optimal point either answer to 2 dp or better or using S.E’s (correct 2 equations for their point + attempt to eliminate one variable.); OR Testing three points correctly and optimal one to 2dp.

2A1 CAO 2 dp or better.

Mark scheme (c)

Scheme	Marks
\(24\dfrac{12}{17} = \dfrac{420}{17} \approx 24.7\) (awrt)	B1
	(1)

Notes

(c)B1 CAO

(Corrected from the printed mark scheme: the scheme prints \(\dfrac{240}{17}\); \(24\dfrac{12}{17} = \dfrac{420}{17}\), as in the list of points above.)

Scheme	Marks
(6,4)	B1
	(1)
	(8 marks)