Maths shelf · Class 12
Price the fair ticket
Raise the ticket price and fewer people come. Lower it and you earn less from each. Where is the best price?
What you will learn: Model takings as a function of price, use the derivative to find where it is greatest, and check the answer against the real-world steps allowed.
What should I know first?
If this feels new, try this first. It is fine to go back.
- The first profitable ticket Class 10
Small test at the end: scroll down to the end of this page for five short questions that check what you have learnt. Tap an answer, then press Check answer. Go to the questions ↓
Jump to: Check yourself · For grown-ups · For teachers
Safety first
Use made-up numbers. Real prices are decided with grown-ups.
Keep it safe:
- Use made-up numbers for practice; real ticket prices and money are decided and handled by grown-ups.
- If you use counters to act out ticket sales, keep them away from babies: nothing goes in the mouth.
You will need
- Graph paperFor the takings curve.
- A calculatorTo fill in the table.
- Paper and a pencilFor the algebra and the derivative.
Time: About 50 minutes.
What do you think?
Tanmay wants ₹100 tickets. Tara wants to keep ₹50. Where do you think the best price is?
Make a prediction and write one reason for it. Then tap the guess closest to yours to see how to test it. It is fine to change your mind after checking.
₹50, more people come
Work out the takings at ₹50 and at ₹60. Did the rise help or hurt?
₹100, more per ticket
How many tickets sell at ₹100? Work out the takings and compare with ₹50.
Somewhere in between
Make a table from ₹50 to ₹100 in ₹10 steps. Where do the takings peak?
Not sure
That is fine. Work out the takings at ₹50, ₹60 and ₹70, and look at the pattern.
Need a hint?
Takings = price × tickets sold. Write both in terms of x, the number of ₹10 rises.
The story
Last year the fair's music show charged ₹50 and sold 100 tickets. The class asked visitors, and found that each ₹10 rise in price would mean about 10 fewer tickets sold. "Charge ₹100 and we'll double the money!" says Tanmay. "Then only 50 people come," says Tara. "That's the same ₹5,000." Dhruv has a feeling the best price is somewhere in between. Dadi, who has priced chai for forty years, just smiles.
The idea
Takings = price × number sold. Here both depend on x, the number of ₹10 rises: T(x) = (50 + 10x)(100 − 10x), for 0 ≤ x ≤ 10.
Expanding: T(x) = 5000 + 500x − 100x². The graph is a parabola that opens downward, so it has a highest point.
At the highest point the curve stops rising and starts falling, so its slope, the derivative, is zero: T′(x) = 500 − 200x = 0 gives x = 2.5.
T″(x) = −200 is negative, which confirms a maximum. Always check the real constraints: whole steps, limits like 0 ≤ x ≤ 10, and the ends of the range.
Worked example
Finding the best price
Find the price that gives the greatest takings.
- T(x) = (50 + 10x)(100 − 10x) = 5000 + 500x − 100x².
- T′(x) = 500 − 200x. Set T′(x) = 0: x = 2.5.
- T″(x) = −200 < 0, so this is a maximum. Price: 50 + 25 = ₹75. Tickets: 100 − 25 = 75. Takings: 75 × 75 = ₹5,625.
- In ₹10 steps only: T(2) = 70 × 80 = ₹5,600 and T(3) = 80 × 70 = ₹5,600. Both ends: T(0) = ₹5,000, T(10) = ₹0.
Best price ₹75 (₹5,625) if any price is allowed; in ₹10 steps, ₹70 or ₹80 (₹5,600).
Try it
Make a table.
For x = 0 to 10, work out the price, tickets sold and takings.
Plot the curve.
Plot takings against x and join the points smoothly. Where is the top?
Write the function.
Write T(x) and expand it. Check it against two rows of your table.
Differentiate.
Find T′(x) and solve T′(x) = 0. Use T″(x) to show it is a maximum.
Respect the rules.
If prices must move in ₹10 steps, what is the best allowed price? Check the ends of the range too.
Advise the class.
Write a short note: what price, why, and what you would watch for if the survey was wrong.
What you make (paper or offline)
Your pricing table
Make your own pricing problem: a starting price, starting sales and how sales change with each rise.
Find the best price with a table, then with the derivative.
Say what real-world limits change your answer.
Make it on paper. Keep it with you. Nothing to upload.
What did we find out?
Read this after you have tried it. Look back at your guess and your reason first.
With x rises of ₹10, the price is 50 + 10x and tickets sold are 100 − 10x, so T = (50 + 10x)(100 − 10x) = 5000 + 500x − 100x².
The derivative dT/dx = 500 − 200x is zero at x = 2.5: a price of ₹75 and the greatest takings, ₹5,625.
If prices can only rise in ₹10 steps, ₹75 is not allowed. Then ₹70 and ₹80 tie at ₹5,600, the best possible.
Check yourself
Five questions about what this page teaches. Choose an answer and press Check answer, or open "See the answer and why". Nothing is saved or sent: your choices stay on this screen and disappear when you leave or reload the page.
See the answer and why
- A. 70 tickets and ₹4,900 Not this one. That uses the price as the number sold. Two rises lose 20 of the 100 sales.
- B. 80 tickets and ₹5,600 Correct. 100 − 20 = 80 tickets. 70 × 80 = 5,600.
- C. 100 tickets and ₹7,000 Not this one. That forgets that fewer people come when the price rises.
- D. 80 tickets and ₹5,000 Not this one. 80 × 70 is 5,600, not 5,000.
See the answer and why
- A. 5000 + 500x − 100x² Correct. 50 × 100 = 5000; 50 × (−10x) + 10x × 100 = 500x; 10x × (−10x) = −100x².
- B. 5000 − 100x² Not this one. This misses the two middle terms, −500x and +1000x.
- C. 5000 + 1500x − 100x² Not this one. 50 × (−10x) is −500x, so the middle terms give 500x, not 1500x.
- D. 5000 + 500x + 100x² Not this one. 10x × (−10x) is −100x², negative.
See the answer and why
- A. 500 − 100x Not this one. The derivative of x² is 2x, so −100x² gives −200x.
- B. 500 − 200x Correct. The 5000 gives 0, 500x gives 500, and −100x² gives −200x.
- C. 500x − 200x² Not this one. Each power of x goes down by one when you differentiate.
- D. 5500 − 200x Not this one. A constant like 5000 has derivative 0.
See the answer and why
- A. x = 2.5, a price of ₹75 Correct. 500 − 200x = 0 gives x = 2.5, so the price is 50 + 25 = ₹75.
- B. x = 0, a price of ₹50 Not this one. At x = 0 the slope is 500, still rising.
- C. x = 2, a price of ₹70 Not this one. T′(2) = 100, so the curve is still rising at x = 2.
- D. x = 5, a price of ₹100 Not this one. That is 500 ÷ 100. Use 500 ÷ 200.
See the answer and why
- A. ₹100, the highest price Not this one. Only 50 come, so takings fall back to ₹5,000.
- B. ₹75, the exact maximum Not this one. ₹75 needs half a ₹10 step, which is not allowed here.
- C. ₹70 or ₹80, both giving ₹5,600 Correct. x must be whole. T(2) = T(3) = 5,600, the highest allowed.
- D. ₹50, because more people come Not this one. More people come, but takings are only ₹5,000.
Think back
Why does the slope of the curve tell you where the top is?
Why must the maths answer (x = 2.5) be checked against the real rules?
Investigate further
- If each ₹10 rise loses 8 tickets instead of 10, what is the best price?
- Add a cost of ₹20 per ticket sold. Maximise profit instead of takings. Does the best price go up or down?
Can you use it here?
A tiffin service sells 60 meals a day at ₹40. Each ₹5 rise loses 5 customers. Find the best price.
Ask someone at home
Ask someone who runs or knows a shop: what happens to sales when they raise prices? Do they ever lower them to sell more?
No one free to join in? That's fine. You can do it by yourself with drawings, and ask a grown-up to help with the materials.
For grown-upsTap a heading to open it. The safety notes are at the top of the page.
Before you start
- Graph paper and a calculator. Keep counters away from babies; nothing in the mouth.
Questions that help
- "What two things change when the price goes up?"
- "Where does the curve stop climbing?"
- "Is that price actually allowed?"
What's really going on
This is a real use of calculus: the best point of a smooth curve is where its rate of change is zero.
The table and graph make the answer believable before the derivative confirms it.
Real decisions have constraints (whole steps, limits). The best allowed answer can differ from the calculus answer.
Other ways to do it
- Use a chai stall: price per cup and cups sold per day.
- Use a tuition batch: fee and number of students.
For teachersLearning aims, classroom ideas and what to look for.
Learning aims
- Form a quadratic model from a real pricing situation.
- Find and justify a maximum using the first and second derivatives.
- Interpret the result within real constraints, including discrete steps and endpoints.
Subject links
- CBSE Class XII Mathematics: Applications of Derivatives (maxima and minima).
- Class 10 "The first profitable ticket" (linear models).
- Economics: demand and revenue.
These links are general. They have not been checked against a specific board syllabus.
In the classroom
- Groups use different survey assumptions and compare best prices.
- Discuss: is the survey reliable? How would you test the "10 fewer per ₹10" assumption?
What to look for
- Builds T(x) from price × quantity.
- Differentiates correctly and checks the second derivative.
- Respects the ₹10-step rule and checks endpoints.