IB Math AI SL/HL Paper 1 Exam Style Practice Questions: Functions-Algebraic Modelling

Piecewise Cost Functions and Mean Price

IB Mathematics HL • Functions • Algebraic Modelling • Piecewise Functions

Lesson Overview

In this lesson, we model the total cost of ordering custom mugs using a piecewise function. The pricing changes after the first 20 mugs, so the total cost must be written carefully in two parts.

This question also links total cost, mean price, and budget constraints. These are common IB modelling skills because students must interpret the wording accurately and build a correct algebraic model before solving.

⭐ Key Concepts

A piecewise function is used when different rules apply in different cases.
If the price per item changes after a certain number, split the total into separate parts.
The mean price is

$\text{mean price}=\frac{\text{total cost}}{\text{number of items}}$

A budget problem often requires solving an inequality and then choosing the greatest whole-number solution.

📘 Clear IB-Style Explanation

For orders of 20 mugs or fewer, every mug costs $p$ dollars. For orders above 20 mugs, the first 20 mugs still cost $p$ dollars each, but every extra mug costs

$p-4$

This means the cost function changes at

$n=20$

so the model must be written carefully.

A common mistake in questions like this is to assume that once the order is above 20, all mugs cost the reduced price. That is not correct. Only the mugs after the first 20 receive the discount.

📌 Worked Example 1 — Forming the Cost Function

A company sells custom mugs.

For an order of 20 mugs or fewer, the price per mug is $p$ dollars.

For larger orders, each mug after the first 20 costs 4 dollars less than $p$ .

Let $C(n)$ be the total cost, in dollars, of ordering $n$ mugs.

(a) Write down an expression, in terms of $p$ , for the total cost of ordering 20 mugs.

(b) Write down an expression for $C(n)$ , in terms of $p$ and $n$ , when $n>20$ .

Solution

(a) Cost of 20 mugs

If each of the 20 mugs costs $p$ dollars, then the total cost is

$20p$

📊 Marks: A1

(b) Cost function for $n>20$

For more than 20 mugs:

the first 20 mugs cost $20p$ ,
the remaining $n-20$ mugs each cost $p-4$ .

$C(n)=20p+(n-20)(p-4)$

This may also be simplified:

$C(n)=20p+np-4n-20p+80$

$C(n)=np-4n+80$

Hence, for $n>20$ ,

$C(n)=20p+(n-20)(p-4)$

or equivalently

$C(n)=np-4n+80$

📊 Marks: M1 A1 A1

📌 Worked Example 2 — Using Mean Price to Find $p$

A business orders 120 mugs. The mean price per mug for this order is $18.50.

(a) Calculate the exact total cost of the order.

(b) Hence find the value of $p$ , correct to 2 decimal places.

Solution

(a) Total cost of 120 mugs

Mean price

$=\frac{\text{total cost}}{\text{number of mugs}}$

$\text{total cost}=120\times18.50$

$=2220$

Therefore, the exact total cost is

$\$2220$

📊 Marks: A1

(b) Find $p$

Since $120>20$ , use

$C(n)=20p+(n-20)(p-4)$

Substitute $n=120$ and $C(120)=2220$ :

$2220=20p+(120-20)(p-4)$

$2220=20p+100(p-4)$

$2220=20p+100p-400$

$2220=120p-400$

$2620=120p$

$p=\frac{2620}{120}$

$p=21.8333\ldots$

$p\approx21.83$

📊 Marks: M1 A1

📌 Worked Example 3 — Using a Budget Constraint

The company’s budget is

$\$2500$

Using your value of $p$ , find the greatest number of mugs the business can order.

Solution

Using

$p\approx21.83$

and for $n>20$ ,

$C(n)=20p+(n-20)(p-4)$

Substitute $p=21.83$ :

$C(n)=20(21.83)+(n-20)(17.83)$

$C(n)=436.6+17.83n-356.6$

$C(n)=17.83n+80$

The budget condition is

$17.83n+80\le2500$

$17.83n\le2420$

$n\le135.726\ldots$

So the greatest whole number of mugs is

$135$

Check:

$C(135)=17.83(135)+80=2487.05$

which is within budget, while

$C(136)=17.83(136)+80=2504.88$

which exceeds the budget.

Therefore, the greatest number of mugs is

$135$

📊 Marks: M1 M1 A1

⚠ Common Mistakes

Assuming that all mugs cost $p-4$ when $n>20$ .
Forgetting that only the mugs after the first 20 are discounted.
Using mean price as though it were total cost.
Not rounding $p$ correctly before using it in the final part.
Giving a non-integer answer for the number of mugs instead of the greatest whole number within budget.

📘 IB Exam Tips

In piecewise cost questions, write the first part and the discounted part separately before simplifying.
Always check whether the order size is above or below the threshold before choosing the formula.
For budget questions, solve the inequality and then test the nearest integers.
Keep the algebra clear when substituting the total cost into the function.
Use the context to guide the final answer: number of mugs must be a whole number.

🧪 Challenge Problem

A company sells custom notebooks.

For an order of 30 notebooks or fewer, the price per notebook is $q$ dollars.

For larger orders, each notebook after the first 30 costs 5 dollars less than $q$ .

A business orders 150 notebooks, and the mean price per notebook is $22.40.

(a) Write down an expression for the total cost of 30 notebooks.

(b) Write down an expression for $C(n)$ when $n>30$ .

(c) Find the exact total cost of 150 notebooks.

(d) Hence find the value of $q$ , correct to 2 decimal places.

(e) If the budget is $3200, find the greatest number of notebooks that can be ordered.

✅ Self-Practice

A business sells custom T-shirts.

For an order of 25 T-shirts or fewer, the price per T-shirt is $k$ dollars.

For larger orders, each T-shirt after the first 25 costs 3 dollars less than $k$ .

A customer orders 100 T-shirts, and the mean price per T-shirt is $16.80.

(a) Write down an expression for the total cost of 25 T-shirts.

(b) Write down an expression for $C(n)$ when $n>25$ .

(c) Find the exact total cost of 100 T-shirts.

(d) Hence find the value of $k$ , correct to 2 decimal places.

(e) If the budget is $1900, find the greatest number of T-shirts that can be ordered.

Show Solutions

Challenge Problem

(a)

$30q$

(b)

$C(n)=30q+(n-30)(q-5)$

(c)

$150\times22.40=3360$

So the exact total cost is

$\$3360$

(d)

$3360=30q+(150-30)(q-5)$

$3360=30q+120(q-5)$

$3360=150q-600$

$3960=150q$

$q=26.4$

$q=26.40$

(e)

For $n>30$ ,

$C(n)=30(26.4)+(n-30)(21.4)$

$C(n)=792+21.4n-642$

$C(n)=21.4n+150$

Budget condition:

$21.4n+150\le3200$

$21.4n\le3050$

$n\le142.52\ldots$

So the greatest number of notebooks is

$142$

Self-Practice

(a)

$25k$

(b)

$C(n)=25k+(n-25)(k-3)$

(c)

$100\times16.80=1680$

So the exact total cost is

$\$1680$

(d)

$1680=25k+(100-25)(k-3)$

$1680=25k+75(k-3)$

$1680=100k-225$

$1905=100k$

$k=19.05$

(e)

For $n>25$ ,

$C(n)=25(19.05)+(n-25)(16.05)$

$C(n)=476.25+16.05n-401.25$

$C(n)=16.05n+75$

Budget condition:

$16.05n+75\le1900$

$16.05n\le1825$

$n\le113.70\ldots$

So the greatest number of T-shirts is

$113$

IB Math AI SL/HL Paper 1 Exam Style Practice Questions: Functions-Algebraic Modelling

Piecewise Cost Functions and Mean Price

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Forming the Cost Function

Solution

📌 Worked Example 2 — Using Mean Price to Find $p$

Solution

📌 Worked Example 3 — Using a Budget Constraint

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

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Piecewise Cost Functions and Mean Price

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Forming the Cost Function

Solution

📌 Worked Example 2 — Using Mean Price to Find

Solution

📌 Worked Example 3 — Using a Budget Constraint

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

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Like this:

Related

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📌 Worked Example 2 — Using Mean Price to Find $p$