Edexcel IALMathematicsGeneral Applications10 Marks

A rectangular box has an open top. The box has length 3x cm, width x cm and height h cm. The volume of the box is 2250 ~cm³.

(a)Show that the surface area A ~cm² of the box is given by A = 3x² + 6000/x.[3]
(b)Find the value of x for which A is a minimum.[5]
(c)Hence find the minimum surface area of the box.[2]

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

(a) M1: Volume V = (3x)(x)(h) = 3x²h. Sets 3x²h = 2250.
(a) A1: Expresses h in terms of x: h = 2250/(3x²) = 750/(x²).
(a) M1: Surface Area A = (3x)(x) + 2(3x)h + 2(x)h (Base + two sides + two ends) (or lists area of 5 faces correctly).
(a) A1: Substitutes their h into A = 3x² + 8xh.
(a) A1: A = 3x² + 8x(750/(x²)) = 3x² + 6000/x. (Must show sufficient working to derive the given expression).
(b) M1: Differentiates A with respect to x: dA/dx. (At least one term correct).
(b) A1: dA/dx = 6x - 6000x⁻² (or 6x - 6000/(x²)).
(b) M1: Sets their dA/dx = 0.
(b) A1: 6x - 6000/(x²) = 0 ⇒ 6x³ = 6000 ⇒ x³ = 1000.
(b) A1: x = 10.
(c) M1: Finds (d²A)/(dx²) (e.g., 6 + 12000x⁻³) and substitutes their x value to show it is a minimum (i.e., $>0$).
(c) A1: Correctly evaluates A = 3(10)² + 6000/10 = 300 + 600 = 900. (Must be minimum area and correct units implied).

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