A container in the shape of a cuboid has a square base of side x and a height of (10-x). It is given that x varies with time, t, where t > 0. The container decreases in volume at a rate which is inversely proportional to t. When t = 1/10, x = 1/2 and the rate of decrease of x is 20/37.

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

1. [B1] Obtain dV/dt = [±]k/t or dV/dt = [±]1/kt
2. [B1] Obtain dV/dx = 20x-3x²
3. [M1] Correct use of chain rule involving k
4. [A1] Obtain dx/dt = [±]-k / (t(20x-3x²)) or equivalent,
5. [A1] Use t = 1/10, x = 1/2 and dx/dt = -20/37 to obtain given answer which must be stated
6. [B1] Separate variables correctly & integrate at least one side correctly
7. [B1] Obtain terms 10x² - x³
8. [B1FT] Obtain term ln t with 'correct' coefficient from their separation of variables, for example a ln t for 1/t
9. [M1] Use t = 1/10, x = 1/2 to evaluate a constant or as limits in a solution containing terms of the form x², x³ and ln t (or ln 2t)
10. [A1] Obtain correct answer in any form, for example 10x² - x³ = (ln t)/2 + 19/8 + (ln 0.1)/2
11. [A1] Obtain answer t = e^(2x³-20x²+19/4) or equivalent