Higher June 2025 Paper 2R Q22
22 The diagram shows a solid cuboid.
Diagram NOT accurately drawn
The volume of the cuboid is \(V\) cm3
Find the maximum value of \(V\)
(5)
| Scheme | Marks |
|---|---|
| \((V =)\;x(x)(15 - 4x)\;(= 15x^2 - 4x^3)\) | M1 |
| \(\left(\dfrac{\mathrm{d}V}{\mathrm{d}x} =\right)\;30x - 12x^2\) | M1ft |
| \(30x - 12x^2 = 0 \Rightarrow x = \ldots\) or \((x =)\;2.5\) oe | M1ft |
| \((V =)\;\text{``}{2.5}\text{''} \times \text{``}{2.5}\text{''} \times (15 - 4 \times \text{``}{2.5}\text{''})\) or \((V =)\;15(\text{``}{2.5}\text{''})^2 - 4(\text{``}{2.5}\text{''})^3\) | M1 |
| Correct answer only scores full marks (unless from obviously incorrect working) Answer: 31.25 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a correct expression for the volume (condone missing \(V = \ldots\))
M1ft: ft dep on a \(V\) of the form \(ax^3 + bx^2\) where \(a, b \ne 0\)
for a two-term derivative with at least one term correct for their \(V\)
eg \(15(2)x\) or \(30x\) or \(-3(4)x^2\) or \(-12x^2\)
M1ft: dep on previous M mark
for equating their 2-term first derivative to 0 and solving to get a value of \(x\) (the value of \(x\) obtained must be greater than 0 and must not be where their second derivative is equal to 0)
or
the correct value of \(x\)
M1: dep on M3
for a full and correct substitution of their value of \(x\)
A1: oe eg \(\dfrac{125}{4}\)
ignore any units on the answer line