IB Mathematics SL 5.6 Local maximum and minimum points AI SL Paper 1- Exam Style Questions- New Syllabus
Question
A company produces and sells electric cars. The company’s profit, P, in thousands of dollars, changes based on the number of cars, x, they produce per month.
The rate of change of their profit from producing x electric cars is modelled by:
\[ \frac{dP}{dx} = -1.6x + 48, \quad x \geq 0. \]
The company makes a profit of 260 (thousand dollars) when they produce 15 electric cars.
(a) Determine an expression for P in terms of x. [4]
(b) The company regularly increases the number of cars it produces. Describe how their profit changes if they increase production to over 30 cars per month and up to 50 cars per month. Justify your answer. [3]
▶️ Answer/Explanation
Markscheme
(a)
Given the rate of change \( \frac{dP}{dx} = -1.6x + 48 \), integrate to find the profit function:
\( P(x) = \int (-1.6x + 48) \, dx = -1.6 \times \frac{x^2}{2} + 48x + c = -0.8x^2 + 48x + c \). M1 A1
Use the condition \( P(15) = 260 \):
Calculate: \( 15^2 = 225 \), \( -0.8 \times 225 = -180 \), \( 48 \times 15 = 720 \).
Substitute: \( 260 = -0.8 \times 15^2 + 48 \times 15 + c = -180 + 720 + c = 540 + c \).
Solve: \( c = 260 – 540 = -280 \). A1
Thus, the expression is \( P(x) = -0.8x^2 + 48x – 280 \). A1
[4 marks]
Given the rate of change \( \frac{dP}{dx} = -1.6x + 48 \), integrate to find the profit function:
\( P(x) = \int (-1.6x + 48) \, dx = -1.6 \times \frac{x^2}{2} + 48x + c = -0.8x^2 + 48x + c \). M1 A1
Use the condition \( P(15) = 260 \):
Calculate: \( 15^2 = 225 \), \( -0.8 \times 225 = -180 \), \( 48 \times 15 = 720 \).
Substitute: \( 260 = -0.8 \times 15^2 + 48 \times 15 + c = -180 + 720 + c = 540 + c \).
Solve: \( c = 260 – 540 = -280 \). A1
Thus, the expression is \( P(x) = -0.8x^2 + 48x – 280 \). A1
[4 marks]
(b)
Analyze the profit function \( P(x) = -0.8x^2 + 48x – 280 \) from \( x = 30 \) to \( x = 50 \).
Method 1 (Derivative Analysis):
The derivative is \( \frac{dP}{dx} = -1.6x + 48 \).
At \( x = 30 \): \( -1.6 \times 30 + 48 = -48 + 48 = 0 \).
For \( x > 30 \), e.g., \( x = 50 \): \( -1.6 \times 50 + 48 = -80 + 48 = -32 \).
Since \( \frac{dP}{dx} < 0 \) for \( x > 30 \), the profit decreases as production increases. M1
Method 2 (Direct Computation):
Calculate: \( P(30) = -0.8 \times 30^2 + 48 \times 30 – 280 = -0.8 \times 900 + 1440 – 280 = -720 + 1440 – 280 = 440 \).
\( P(50) = -0.8 \times 50^2 + 48 \times 50 – 280 = -0.8 \times 2500 + 2400 – 280 = -2000 + 2400 – 280 = 120 \).
Profit decreases from 440 to 120 thousand dollars, a drop of 320 thousand dollars. A1
Method 3 (Integration):
The change in profit is \( \int_{30}^{50} (-1.6x + 48) \, dx = [-0.8x^2 + 48x]_{30}^{50} \).
Evaluate: \( (-0.8 \times 50^2 + 48 \times 50) – (-0.8 \times 30^2 + 48 \times 30) = (120) – (440) = -320 \).
The negative value confirms a decrease of 320 thousand dollars.
Thus, the profit decreases with each additional car produced from 30 to 50 cars per month. A1
[3 marks]
Analyze the profit function \( P(x) = -0.8x^2 + 48x – 280 \) from \( x = 30 \) to \( x = 50 \).
Method 1 (Derivative Analysis):
The derivative is \( \frac{dP}{dx} = -1.6x + 48 \).
At \( x = 30 \): \( -1.6 \times 30 + 48 = -48 + 48 = 0 \).
For \( x > 30 \), e.g., \( x = 50 \): \( -1.6 \times 50 + 48 = -80 + 48 = -32 \).
Since \( \frac{dP}{dx} < 0 \) for \( x > 30 \), the profit decreases as production increases. M1
Method 2 (Direct Computation):
Calculate: \( P(30) = -0.8 \times 30^2 + 48 \times 30 – 280 = -0.8 \times 900 + 1440 – 280 = -720 + 1440 – 280 = 440 \).
\( P(50) = -0.8 \times 50^2 + 48 \times 50 – 280 = -0.8 \times 2500 + 2400 – 280 = -2000 + 2400 – 280 = 120 \).
Profit decreases from 440 to 120 thousand dollars, a drop of 320 thousand dollars. A1
Method 3 (Integration):
The change in profit is \( \int_{30}^{50} (-1.6x + 48) \, dx = [-0.8x^2 + 48x]_{30}^{50} \).
Evaluate: \( (-0.8 \times 50^2 + 48 \times 50) – (-0.8 \times 30^2 + 48 \times 30) = (120) – (440) = -320 \).
The negative value confirms a decrease of 320 thousand dollars.
Thus, the profit decreases with each additional car produced from 30 to 50 cars per month. A1
[3 marks]
Total Marks: 7