# Question 9:

## Part (a)

| Answer | Mark | Guidance |
|--------|------|----------|
| $x \geqslant 100$ | B1 | At least 100 soft pillows |
| $y \geqslant 200$ | B1 | At least 200 medium pillows |
| $x + y + z \geqslant 400$ | B1 | At least 400 pillows in total |
| $4x + 3y + 4z \leqslant 1800$ | B1 | Spend no more than £1800 |
| $y \geqslant \frac{2}{5}(x + y + z)$ or equivalent e.g. $3y \geqslant 2x + 2z$ or $2x - 3y + 2z \leqslant 0$ | B1 | At least 40% must be medium; allow $y \geqslant 0.4(x+y+z)$ |
| **3 of the above 5 correctly stated** | **3 marks total** | Award B1 for each correct inequality up to 3 marks |

## Part (b)(i)

| Answer | Mark | Guidance |
|--------|------|----------|
| Substituting $x = 2z$ (i.e. $z = \frac{x}{2}$) into $x + y + z \geqslant 400$ gives $\frac{3x}{2} + y \geqslant 400$, so $3x + 2y \geqslant 800$ | M1 | Show substitution of $z = \frac{x}{2}$ into total constraint |
| Substituting $z = \frac{x}{2}$ into $4x + 3y + 4z \leqslant 1800$ gives $4x + 3y + 2x \leqslant 1800$, so $2x + y \leqslant 600$ | A1 | Correct derivation shown |
| Substituting $z = \frac{x}{2}$ into $3y \geqslant 2x + 2z$ gives $3y \geqslant 2x + x$, so $y \geqslant x$ | A1 | All three shown correctly |

## Part (b)(ii)

| Answer | Mark | Guidance |
|--------|------|----------|
| Line $3x + 2y = 800$ drawn correctly | B1 | Passes through e.g. $(0, 400)$ and $(200, 100)$ or equivalent correct points |
| Line $2x + y = 600$ drawn correctly | B1 | Passes through e.g. $(0, 600)$ and $(300, 0)$ |
| Line $y = x$ drawn correctly | B1 | Diagonal through origin at 45° |
| Line $x = 100$ drawn (from part a) | B1 | Vertical line at $x = 100$ |
| Line $y = 200$ drawn (from part a) | B1 | Horizontal line at $y = 200$ |
| Feasible region correctly indicated/shaded | B1 | Must be consistent with all lines drawn; award only if at least 3 lines correct |

## Part (b)(iii)

| Answer | Mark | Guidance |
|--------|------|----------|
| Objective: maximise $x + y + z = x + y + \frac{x}{2} = \frac{3x}{2} + y$ | M1 | Correct identification of objective function to maximise |
| Maximum at vertex of feasible region, e.g. intersection of $2x + y = 600$ and $y = x$: $x = 200, y = 200$, total $= \frac{3(200)}{2} + 200 = 500$ | A1 | **Maximum total = 500 pillows** |

## Part (b)(iv)

| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 200$ soft, $y = 200$ medium, $z = 100$ firm (since $z = \frac{x}{2} = 100$) | B1 | Follow through from (b)(iii); all three values required |