IB Mathematics SL 5.6 Local maximum and minimum points AI HL Paper 2- Exam Style Questions- New Syllabus
Question
A store employs the following model to predict \( n \), the number of smoothies sold daily, based on \( x \), the cost of one smoothie in pesos.
\( n = \frac{40000}{x^2} \)
The highest number of smoothies the store can prepare in a day is 400.
(a) Determine the highest price they could set for a smoothie to sell all 400 in a single day.
(b) On a day when the store sells smoothies at 50 pesos each, use the model to find
(i) the quantity of smoothies sold.
(ii) the total revenue from the smoothies sold.
The expense of producing each smoothie is 20 pesos. The daily profit (\( P \)) is calculated as the total revenue from smoothie sales minus the production costs.
(c) (i) Demonstrate that, according to the model,
\( P = \frac{40000}{x} – \frac{800000}{x^2}. \)
(ii) Calculate \( \frac{dP}{dx}. \)
(iii) Identify the value of \( x \) where \( \frac{dP}{dx}=0. \)
(iv) Determine the number of smoothies sold when the profit reaches its maximum.
▶️ Answer/Explanation
Model: \( n = \frac{40000}{x^2} \); where \( n \) is the number of smoothies sold per day and \( x \) is the price of a smoothie (in pesos); Maximum capacity = 400 smoothies/day.
(a) Maximum price to sell 400 smoothies:
\( \frac{40000}{x^2} = 400 \); \( x^2 = \frac{40000}{400} = 100 \); \( x = 10 \)
Answer: \( x = 10 \) pesos.
(b)(i) At \( x = 50 \) pesos, number of smoothies:
\( n = \frac{40000}{50^2} \); \( = \frac{40000}{2500} \); \( = 16 \)
Answer: 16 smoothies.
(b)(ii) Total income (revenue):
\( R = 50 × 16 \); \( = 800 \)
Answer: 800 pesos.
(c)(i) Profit function:
- Revenue: \( R = x × n = x × \frac{40000}{x^2} \); \( = \frac{40000}{x} \)
- Cost: \( C = 20n = 20 × \frac{40000}{x^2} \); \( = \frac{800000}{x^2} \)
\( P = R – C = \frac{40000}{x} – \frac{800000}{x^2} \)
Answer: \( P = \frac{40000}{x} – \frac{800000}{x^2} \)
(c)(ii) Differentiate:
\( \frac{dP}{dx} = \frac{d}{dx}\!\left(40000x^{-1} – 800000x^{-2}\right) \); \( \frac{dP}{dx} = -40000x^{-2} + 1600000x^{-3} \)
Answer: \( \frac{dP}{dx} = -\frac{40000}{x^2} + \frac{1600000}{x^3} \)
(c)(iii) Solve \( \frac{dP}{dx} = 0 \):
\( -\frac{40000}{x^2} + \frac{1600000}{x^3} = 0 \); Multiply through by \( x^3 \); \( -40000x + 1600000 = 0 \); \( x = 40 \)
Second derivative test:
\( \frac{d^2P}{dx^2} = \frac{80000}{x^3} – \frac{4800000}{x^4} \); At \( x = 40 \); \( \frac{d^2P}{dx^2} = 1.25 – 1.875 \); \( = -0.625 < 0 \)
Conclusion: Profit is maximized at \( x = 40 \) pesos.
(c)(iv) Number of smoothies at max profit:
\( n = \frac{40000}{40^2} \); \( = \frac{40000}{1600} \); \( = 25 \)
Answer: 25 smoothies; (Max profit value: \( P(40) = 1000 – 500 = 500 \) pesos)