IB Mathematics SL 5.1 Derivative interpreted as gradient function AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Myke operates a business specializing in the production and sale of curry powder. The marginal profit, \( P \), in Mauritian rupees (MUR), relative to the quantity \( x \) kilograms (kg) produced, is given by the following derivative:
\( \frac{dP}{dx} = -10x + 460, \quad x \geq 0. \)
It is known that the company earns a total profit of 3300 MUR when the production level is 10 kg.
(a) Determine an equation for the total profit, \( P \), as a function of \( x \).
Myke intends to expand her production from 25 kg to 50 kg.
(b) Calculate the resulting increase in the company’s profit.
Most-appropriate topic codes (IB Mathematics AI SL 2025):
• SL 5.5: Anti-differentiation and definite integrals — parts (a), (b)
• SL 5.7: Optimization in context — part (b) interpretation
• SL 5.1: Derivative as rate of change — part (a) setup
• SL 5.7: Optimization in context — part (b) interpretation
• SL 5.1: Derivative as rate of change — part (a) setup
▶️ Answer/Explanation
(a)
Integrate \(\frac{dP}{dx}\):
\(P = \int (-10x + 460) \, dx = -5x^2 + 460x + c.\)
Use the condition \(P = 3300\) when \(x = 10\):
\(3300 = -5(10)^2 + 460(10) + c \implies 3300 = -500 + 4600 + c \implies c = -800.\)
Thus,
\(P(x) = -5x^2 + 460x – 800.\)
\(\boxed{P(x) = -5x^2 + 460x – 800}\)
(b)
Increase in profit = \(P(50) – P(25)\):
\(P(50) = -5(50)^2 + 460(50) – 800 = 14200,\)
\(P(25) = -5(25)^2 + 460(25) – 800 = 6675.\)
Increase = \(14200 – 6675 = 7525\) MUR.
Alternatively, using the integral of the rate:
\(\int_{25}^{50} (-10x + 460) \, dx = \left[ -5x^2 + 460x \right]_{25}^{50} = 7525.\)
\(\boxed{7525 \text{ MUR}}\)