Problem 6: Problem Sum - Matrix multiplication to determine total delivery costs for coffee beans.

Let's solve the problem step by step.

---

Problem Summary:

A supplier supplies coffee beans to shops A, B, and C in bags of 5 kg, 10 kg, and 15 kg. The quantities delivered are given in a table:

| | 5 kg | 10 kg | 15 kg |
|----------|------|-------|-------|
| Shop A | 50 | 100 | 30 |
| Shop B | 60 | 150 | 40 |
| Shop C | 80 | 200 | 70 |

Cost prices:
- 5 kg bag: $10.60
- 10 kg bag: $17.20
- 15 kg bag: $22.50

---

Part (a):

We need to define two matrices:
- Matrix P: represents the number of bags delivered to each shop.
- Matrix Q: represents the cost per bag for each size.

Then compute the product PQ, which gives the total cost to each shop.

#### Step 1: Define Matrix P

Matrix P is a 3×3 matrix where rows represent shops and columns represent bag sizes.

$$
P = \begin{bmatrix}
50 & 100 & 30 \\
60 & 150 & 40 \\
80 & 200 & 70 \\
\end{bmatrix}
$$

Rows: Shops A, B, C
Columns: 5kg, 10kg, 15kg

#### Step 2: Define Matrix Q

Matrix Q is a 3×1 column matrix representing the cost per bag for each size.

$$
Q = \begin{bmatrix}
10.60 \\
17.20 \\
22.50 \\
\end{bmatrix}
$$

Note: Since we're multiplying P × Q, the number of columns in P must match the number of rows in Q, which it does (both have 3).

#### Step 3: Compute PQ

Each element in the resulting matrix (which will be 3×1) is the total cost for that shop.

$$
PQ = \begin{bmatrix}
50 & 100 & 30 \\
60 & 150 & 40 \\
80 & 200 & 70 \\
\end{bmatrix}
\times
\begin{bmatrix}
10.60 \\
17.20 \\
22.50 \\
\end{bmatrix}
$$

Now compute each row:

---

Shop A:
$$
50 \times 10.60 + 100 \times 17.20 + 30 \times 22.50
= 530 + 1720 + 675 = 2925
$$

Shop B:
$$
60 \times 10.60 + 150 \times 17.20 + 40 \times 22.50
= 636 + 2580 + 900 = 4116
$$

Shop C:
$$
80 \times 10.60 + 200 \times 17.20 + 70 \times 22.50
= 848 + 3440 + 1575 = 6863
$$

So,

$$
PQ = \begin{bmatrix}
2925 \\
4116 \\
6863 \\
\end{bmatrix}
$$

This means:
- Total cost to Shop A: $2925
- Total cost to Shop B: $4116
- Total cost to Shop C: $6863

✔ This answers part (a).

---

Part (b):

In June, the deliveries were:
- Shop A: 15 deliveries
- Shop B: 10 deliveries
- Shop C: 5 deliveries

We are to use matrix multiplication to find the total number of bags of each size delivered in June.

We assume that each delivery has the same quantity as in the original table.

So, we can think of this as:
- Multiply the original delivery matrix P by a column vector of delivery counts.

Let’s define a new matrix or vector:

Let D be a 3×1 vector:

$$
D = \begin{bmatrix}
15 \\
10 \\
5 \\
\end{bmatrix}
$$

Now, multiply P × D → this gives total number of bags of each type delivered in June.

But note: P is 3×3 (shops × bag sizes), and D is 3×1 (delivery counts per shop). So:

$$
\text{Total bags} = P \times D
$$

Wait! Actually, we need to multiply the delivery counts with the bag quantities per shop.

But here's the key: Each delivery to a shop includes the same mix of bags.

So if Shop A gets 15 deliveries, and each delivery includes 50 bags of 5kg, etc., then total bags for Shop A = 15 × [50, 100, 30]

Similarly for others.

So we can write:

$$
\text{Total bags} = P \times D
$$

But P is 3×3, D is 3×1 → result is 3×1, but we want total per bag size.

Wait — actually, we want to sum over all shops.

So better to do:

Let’s define a row vector of delivery counts:
$$
R = \begin{bmatrix}
15 & 10 & 5 \\
\end{bmatrix}
$$

And multiply it by P (which is 3×3):

$$
R \times P = \begin{bmatrix}
15 & 10 & 5 \\
\end{bmatrix}
\times
\begin{bmatrix}
50 & 100 & 30 \\
60 & 150 & 40 \\
80 & 200 & 70 \\
\end{bmatrix}
$$

Result will be a 1×3 row vector giving total number of 5kg, 10kg, and 15kg bags delivered in June.

Let’s compute:

Total 5kg bags:
$$
15 \times 50 + 10 \times 60 + 5 \times 80 = 750 + 600 + 400 = 1750
$$

Total 10kg bags:
$$
15 \times 100 + 10 \times 150 + 5 \times 200 = 1500 + 1500 + 1000 = 4000
$$

Total 15kg bags:
$$
15 \times 30 + 10 \times 40 + 5 \times 70 = 450 + 400 + 350 = 1200
$$

So total bags delivered in June:

- 5 kg: 1750
- 10 kg: 4000
- 15 kg: 1200

✔ Total number of bags = 1750 + 4000 + 1200 = 6950

---

✔ Final Answers:

#### (a)
Matrices:
$$
P = \begin{bmatrix}
50 & 100 & 30 \\
60 & 150 & 40 \\
80 & 200 & 70 \\
\end{bmatrix}, \quad
Q = \begin{bmatrix}
10.60 \\
17.20 \\
22.50 \\
\end{bmatrix}
$$

Product:
$$
PQ = \begin{bmatrix}
2925 \\
4116 \\
6863 \\
\end{bmatrix}
$$
→ Total cost to each shop.

#### (b)
Using matrix multiplication:
$$
R \times P = \begin{bmatrix}
15 & 10 & 5 \\
\end{bmatrix}
\times
\begin{bmatrix}
50 & 100 & 30 \\
60 & 150 & 40 \\
80 & 200 & 70 \\
\end{bmatrix}
=
\begin{bmatrix}
1750 & 4000 & 1200 \\
\end{bmatrix}
$$

So:
- Total 5 kg bags: 1750
- Total 10 kg bags: 4000
- Total 15 kg bags: 1200

→ Total number of bags delivered in June: 6950

---

🔍 Note on Your Handwritten Work:

You wrote:
> Shop A: 50×10.60 + 100×17.20 + 30×22.50 = ?

That's correct for Shop A’s total cost, which is $2925 — matches our calculation.

But you didn't complete the rest. Also, this is only part (a). For part (b), you need to use matrix multiplication to scale the deliveries.

---

✔ Summary:

- Use P for quantities, Q for costs → PQ gives total cost per shop.
- Use R × P (with R = [15,10,5]) to get total bags per size in June.
- Total bags in June: 6950

Let me know if you'd like a diagram or visual representation!