4MA0 Higher June 2014 Paper 3 Q14

EdexcelHigherOld spec8 marksDifferentiation Forming Equations

14 A farmer has 180 metres of fencing.
With the 180 metres of fencing, he makes an enclosure divided into eight equal, rectangular pens.
The fencing is used for the perimeter of each pen.

Rectangle divided into 2 rows of 4 equal pens; each pen has length x and width y

The length of each pen is \(x\) metres and the width of each pen is \(y\) metres.

(a)

(i) Show that \(y = 18 - 1.2x\)

The total area of the enclosure is \(A\) m².

(ii) Show that \(A = 144x - 9.6x^2\)

(3)

(b) Find \(\dfrac{\mathrm{d}A}{\mathrm{d}x}\) (2)

Mark scheme (a)

Scheme	Marks
(i) \(12x + 10y = 180\)	B1
(ii) \((A =)\ 4x \times 2y\)	M1
\((A =)\ 4x \times 2(18 - 1.2x)\) Answer: proceed to \(A = 144x - 9.6x^2\)	A1
	(3)

Notes

B1: Accept \(12x = 180 - 10y\) or \(10y = 180 - 12x\)

M1: \(4x \times 2y\) or \(8xy\) oe

A1: \(4x \times 2(18 - 1.2x)\) or \(8x(18 - 1.2x)\) or \(4x(36 - 2.4x)\) oe AND proceeding correctly to \(A = 144x - 9.6x^2\)

Scheme	Marks
\((\mathrm{d}A/\mathrm{d}x =)\ 144 - 19.2x\)	B2
	(2)

Mark scheme (c)

Scheme	Marks
\(\text{“}144 - 19.2x\text{”} = 0\)	M1 ft
\(x = 7.5\ (y = 9)\) \((A =)\ 144 \times \text{“}7.5\text{”} - 9.6 \times \text{“}7.5\text{”}^2\) or \((A =)\ 8 \times \text{“}7.5\text{”} \times \text{“}9\text{”}\)	M1 dep
540	A1
	(3)
	(8 marks)

Notes

M1 ft: Must be a 2 part linear expression