Case Study II

Raman Lal runs a stationery shop in Pune. The analysis of his sales, expenditures and profits showed that for x number of notebooks sold, the weekly profit (in ₹) was P(x) = - 2x² + 88x - 680. Raman Lal found that:

• He has a loss if he does not sell any notebook in a week.
• There is no profit no loss for a certain value x₀ of x.
• The profit goes on increasing with an increase in x i.e. the number of notebooks sold. But he gets a maximum profit at a sale of 22 notebooks in a week.

Now answer the following questions :

1. What will be Raman Lal’s profit if he sold 20 notebooks in a week?

1. ₹ 144
2. ₹ 280
3. ₹ 340
4. ₹ 560

2. What is the maximum profit that Raman Lal can earn in a week?

1. ₹ 144
2. ₹ 288
3. ₹ 340
4. ₹ 680

3. What is Raman Lal’s loss if he does not sell any notebooks in a particular week?

1. ₹ 0
2. ₹ 340
3. ₹ 680
4. ₹ 960

4. Write a quadratic equation for the condition when Raman Lal does not have any profit or loss during a week.

1. 2x² - 44x + 340 = 0
2. x² + 44x - 340 = 0
3. x² - 88x + 340 = 0
4. x² - 44x + 340 = 0

5. What is the minimum number of notebooks x₀ that Raman Lal should sell in a week so that he does not incur any loss?

1. 0
2. 10
3. 11
4. 12

1. Given,

⇒ P(x) = -2x² + 88x - 680

Books sold by Raman lal is 20.Therefore x = 20

⇒ P(20) = −2(20)² + 88(20) − 680

⇒ P(20) = −2(400) + 1760 − 680

⇒ P(20) = −800 + 1080

⇒ P(20) = 280.

Hence, option (b) is the correct option.

2. Given,

⇒ P(x) = - 2x² + 88x - 680

Maximum profit occurs when 22 books are sold, thus x = 22.

⇒ P(22) = −2(22)² + 88(22) − 680

= −2(484) + 1936 − 680

= −968 + 1256

= ₹ 288.

Hence, option (b) is the correct option.

3. Given,

⇒ Number of notebooks sold in a particular week = 0

Thus, x = 0

The weekly profit P(x) = -2x² + 88x - 680.

Raman Lal’s loss if no notebooks sold in a week,

⇒ P(0) = -2(0)² + 88(0) - 680

= −680

= loss of ₹ 680.

Hence, option (c) is the correct option.

4. Given,

⇒ P(x) = -2x² + 88x - 680

If there is no profit/loss, then profit = ₹ 0.

⇒ P(x) = 0

⇒ −2x² + 88x − 680 = 0

⇒ −2(x² - 44x + 340) = 0

⇒ x² − 44x + 340 = 0.

Hence, option (d) is the correct option.

5. When Raman lal has no profit/loss we get the equation x² − 44x + 340 = 0.

⇒ x² − 44x + 340 = 0

⇒ x² − 10x - 34x + 340 = 0

⇒ x(x − 10) - 34(x - 10) = 0

⇒ (x − 10)(x - 34) = 0

⇒ (x − 10) = 0 or (x - 34) = 0      [Using Zero-product rule]

⇒ x = 10 or x = 34.

The minimum value of x with no loss is x = 10. Raman lal has to sell 10 books so that he does not incur any loss.

Hence, option (b) is the correct option.