June 2025 Paper 1 Q18

18 A 60 cm piece of wire is bent to form three sides of a rectangle.

It is then attached to a fixed rod as shown in the diagram.

(a)

The area of the rectangle is \(A\) cm\(^2\)

Show that \(\quad A = x(60 - 2x)\) [2 marks]

(b) Work out the maximum value of \(A\) as \(x\) varies. [4 marks]

Mark scheme (a)

Answer	Mark	Comments
\(2x + y = 60\) or \(y = 60 - 2x\) seen	M1	oe
\(A = xy = x(60 - 2x)\) seen	A1	must be clear that the answer has come from an understanding that the area is \(x\) multiplied by \(y\) eg \(A =\) length \(\times\) width

Consider working in part (b) if not contradictory
Condone \(A = x \times y = x \times (60 - 2x)\)

Mark scheme (b)

Answer	Mark	Comments
Alternative method 1: differentiation
\(60x - 2x^2\)	M1	expanding the brackets
\(\left(\dfrac{\mathrm{d}A}{\mathrm{d}x} =\right) 60 - 4x\)	M1dep	oe condone \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\)
\(x = 15\)	A1ft	dependent on first M mark \(x\) value from their 2-term derivative
450	A1
Alternative method 2: midpoint of roots
\(x(60 - 2x) = 0\) or \(2x(30 - x) = 0\)	M1	only award if it is clearly an attempt to find the midpoint of the roots
\(x = 0\) and \(x = 30\)	M1dep	with M1 awarded
\(x = 15\)	A1ft	dependent on first M mark correct midpoint from their roots
450	A1
Alternative method 3: completes the square
\(60x - 2x^2\) or \(-2(x^2 - 30x)\)	M1
\(-2(x - 15)^2 \ldots\ldots\)	M1dep
\(x = 15\)	A1ft	dependent on first M mark correct value from their bracket
450	A1	implies A1ft

Consider working in part (a) and on diagram if not contradictory
Answer only without any working gains full marks (students may know that the maximum Area is when the length is twice the width)	M2A2
450 from Trial and Improvement (must be on answer line)	M2A2
Incorrect answer from Trial and Improvement (but M marks could have been awarded before T&I started)	M0A0
On Alt 1 condone \(A = 60 - 4x\) and \(y = 60 - 4x\) in Alt 1