D1 January 2011 Q6
6.

The graph in Figure 6 is being used to solve a linear programming problem.
Two of the constraints have been drawn on the graph and the rejected regions shaded out.
Two further constraints are
\[\begin{aligned}x + y &\geqslant 30\\ \text{and}\quad 5x + 8y &\leqslant 400\end{aligned}\]The objective is to
\[\text{minimise }\ 15x + 10y\]| Scheme | Marks |
|---|---|
![]() | |
| \(4y \geqslant x\) o.e. | B1 B1 |
| \(2y \leqslant x + 30\) o.e | B1 B1 |
| (4) |
Notes
1B1: ratio of coefficients correct (i.e. equation of line correct)
2B1: inequality correct way round.(\(ay \geqslant bx\) o.e.)
3B1: ratio of coefficients correct (i.e equation of line correct)
4B1: inequality correct way round.
| Scheme | Marks |
|---|---|
| \(x + y = 30\) and \(5x + 8y = 400\) added to the graph | B1, B1 |
| shading correct | B1ft |
| R correct | B1 |
| (4) |
Notes
1B1: \(x + y = 30\) drawn cao
2B1: \(5x + 8y = 400\) drawn cao
3B1ft: shading correct or implied from lines with negative gradient.
4B1: cao
| Scheme | Marks |
|---|---|
| Profit line attempted | M1 |
| Correct profit line | A1 |
| (10, 20) | B1 |
| (3) | |
| (11 marks) |
Notes
1M1: Profit line – intersecting both axes. Minimum (2,0) to (0,3). Accept reciprocal gradient here.
1A1: a correct line
2A1=1B1: cao (e.g not ‘10x + 20y’)
