Question 6 - A-Level Maths

AQA D1 2009 June — Question 6 21 marks

Exam Board	AQA
Module	D1 (Decision Mathematics 1)
Year	2009
Session	June
Marks	21
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear Programming
Type	Three-variable constraint reduction
Difficulty	Moderate -0.3 This is a standard D1 linear programming question requiring constraint formulation and graphical solution. Part (a) involves straightforward translation of word problems into inequalities with basic simplification. Part (b) reduces to a 2D problem through substitution (z=x), then uses routine graphical methods to find optimal solutions. While multi-step, it follows the standard D1 template without requiring novel insight or complex reasoning beyond textbook techniques.
Spec	7.06a LP formulation: variables, constraints, objective function 7.06d Graphical solution: feasible region, two variables

6 Each day, a factory makes three types of widget: basic, standard and luxury. The widgets produced need three different components: type $A$, type $B$ and type $C$. Basic widgets need 6 components of type $A , 6$ components of type $B$ and 12 components of type $C$.
Standard widgets need 4 components of type $A , 3$ components of type $B$ and 18 components of type $C$.
Luxury widgets need 2 components of type $A , 9$ components of type $B$ and 6 components of type $C$.
Each day, there are 240 components of type $A$ available, 300 of type $B$ and 900 of type $C$.
Each day, the factory must use at least twice as many components of type $C$ as type $B$.
Each day, the factory makes $x$ basic widgets, $y$ standard widgets and $z$ luxury widgets.

In addition to $x \geqslant 0 , y \geqslant 0$ and $z \geqslant 0$, find four inequalities in $x , y$ and $z$ that model the above constraints, simplifying each inequality.
Each day, the factory makes the maximum possible number of widgets. On a particular day, the factory must make the same number of luxury widgets as basic widgets.
1. Show that your answers in part (a) become $$2 x + y \leqslant 60 , \quad 5 x + y \leqslant 100 , \quad x + y \leqslant 50 , \quad y \geqslant x$$
2. Using the axes opposite, draw a suitable diagram to enable the problem to be solved graphically, indicating the feasible region.
3. Find the total number of widgets made on that day.
4. Find all possible combinations of the number of each type of widget made that correspond to this maximum number.

Show mark scheme Show mark scheme source

Question 6:

Part (a):

Answer	Marks	Guidance
Answer	Marks	Guidance
Type A: $6x + 4y + 2z \leq 240 \Rightarrow 3x + 2y + z \leq 120$	B1	Must be simplified
Type B: $6x + 3y + 9z \leq 300 \Rightarrow 2x + y + 3z \leq 100$	B1	Must be simplified
Type C: $12x + 18y + 6z \leq 900 \Rightarrow 2x + 3y + z \leq 150$	B1	Must be simplified
C ≥ 2B: $12x + 18y + 6z \geq 2(6x + 3y + 9z) \Rightarrow 12z \leq 6y \Rightarrow y \geq 2z$	B1	Must be simplified
Each inequality correct unsimplified	M1
Each correct simplified	A1 each	Up to 4 A1s

Part (b)(i):

Answer	Marks	Guidance
Answer	Marks	Guidance
Substituting $z = x$ into inequalities	M1
$3x + 2y + x \leq 120 \Rightarrow 2x + y \leq 60$ ✓	A1
$2x + y + 3x \leq 100 \Rightarrow 5x + y \leq 100$ ✓	A1
$x + y \leq 50$ and $y \geq x$ shown	A1

Part (b)(ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Line $2x + y = 60$ correctly drawn	B1
Line $5x + y = 100$ correctly drawn	B1
Line $x + y = 50$ correctly drawn	B1
Line $y = x$ correctly drawn	B1
Feasible region correctly identified and labelled	B1

Part (b)(iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Objective is to maximise $x + y + z = 2x + y$ (since $z = x$)	M1
Maximum total = 50 widgets	A1

Part (b)(iv):

Answer	Marks	Guidance
Answer	Marks	Guidance
Testing vertices of feasible region	M1
$(10, 30)$: $x=10, y=30, z=10$ giving total 50	A1
$(0, 50)$: $x=0, y=50, z=0$ giving total 50	A1
All valid combinations on line segment between these vertices stated	A1	Accept integer solutions along $x+y=50, y\geq x$

## Question 6:

### Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Type A: $6x + 4y + 2z \leq 240 \Rightarrow 3x + 2y + z \leq 120$ | B1 | Must be simplified |
| Type B: $6x + 3y + 9z \leq 300 \Rightarrow 2x + y + 3z \leq 100$ | B1 | Must be simplified |
| Type C: $12x + 18y + 6z \leq 900 \Rightarrow 2x + 3y + z \leq 150$ | B1 | Must be simplified |
| C ≥ 2B: $12x + 18y + 6z \geq 2(6x + 3y + 9z) \Rightarrow 12z \leq 6y \Rightarrow y \geq 2z$ | B1 | Must be simplified |
| Each inequality correct unsimplified | M1 | |
| Each correct simplified | A1 each | Up to 4 A1s |

### Part (b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substituting $z = x$ into inequalities | M1 | |
| $3x + 2y + x \leq 120 \Rightarrow 2x + y \leq 60$ ✓ | A1 | |
| $2x + y + 3x \leq 100 \Rightarrow 5x + y \leq 100$ ✓ | A1 | |
| $x + y \leq 50$ and $y \geq x$ shown | A1 | |

### Part (b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Line $2x + y = 60$ correctly drawn | B1 | |
| Line $5x + y = 100$ correctly drawn | B1 | |
| Line $x + y = 50$ correctly drawn | B1 | |
| Line $y = x$ correctly drawn | B1 | |
| Feasible region correctly identified and labelled | B1 | |

### Part (b)(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Objective is to maximise $x + y + z = 2x + y$ (since $z = x$) | M1 | |
| Maximum total = **50** widgets | A1 | |

### Part (b)(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Testing vertices of feasible region | M1 | |
| $(10, 30)$: $x=10, y=30, z=10$ giving total 50 | A1 | |
| $(0, 50)$: $x=0, y=50, z=0$ giving total 50 | A1 | |
| All valid combinations on line segment between these vertices stated | A1 | Accept integer solutions along $x+y=50, y\geq x$ |

---

Show LaTeX source

6 Each day, a factory makes three types of widget: basic, standard and luxury. The widgets produced need three different components: type $A$, type $B$ and type $C$.

Basic widgets need 6 components of type $A , 6$ components of type $B$ and 12 components of type $C$.\\
Standard widgets need 4 components of type $A , 3$ components of type $B$ and 18 components of type $C$.\\
Luxury widgets need 2 components of type $A , 9$ components of type $B$ and 6 components of type $C$.\\
Each day, there are 240 components of type $A$ available, 300 of type $B$ and 900 of type $C$.\\
Each day, the factory must use at least twice as many components of type $C$ as type $B$.\\
Each day, the factory makes $x$ basic widgets, $y$ standard widgets and $z$ luxury widgets.
\begin{enumerate}[label=(\alph*)]
\item In addition to $x \geqslant 0 , y \geqslant 0$ and $z \geqslant 0$, find four inequalities in $x , y$ and $z$ that model the above constraints, simplifying each inequality.
\item Each day, the factory makes the maximum possible number of widgets. On a particular day, the factory must make the same number of luxury widgets as basic widgets.
\begin{enumerate}[label=(\roman*)]
\item Show that your answers in part (a) become

$$2 x + y \leqslant 60 , \quad 5 x + y \leqslant 100 , \quad x + y \leqslant 50 , \quad y \geqslant x$$
\item Using the axes opposite, draw a suitable diagram to enable the problem to be solved graphically, indicating the feasible region.
\item Find the total number of widgets made on that day.
\item Find all possible combinations of the number of each type of widget made that correspond to this maximum number.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA D1 2009 Q6 [21]}}

This paper (7 questions)

View full paper

Q1 7 Q2 6 Q3 10 Q4 13 Q5 10 Q6 21 Q7 8