June 2022 Paper 2 Q7

7 The curve \(y = 15 - x^2\) and the isosceles triangle \(OPQ\) are shown on the diagram below.

The n-shaped curve y = 15 − x² crossing the x-axis either side of O; P and Q lie on the curve at the same height, Q above the point q on the positive x-axis; the shaded triangle OPQ has its vertex at the origin O

Vertices \(P\) and \(Q\) lie on the curve such that \(Q\) lies vertically above some point \((q, 0)\)

The line \(PQ\) is parallel to the \(x\)-axis.

(a) Show that the area, \(A\), of the triangle \(OPQ\) is given by\[A = 15q - q^3 \quad \text{for } 0 \lt q \lt c\]where \(c\) is a constant to be found. [3 marks]

(b) Find the exact maximum area of triangle \(OPQ\).

Fully justify your answer. [6 marks]

Triangle with horizontal top side of length 2q and height 15 − q², split down the middle — Mark scheme (a)

Scheme	Marks	AO
Identifies the height of the triangle or rectangle as \(15 - q^2\) PI by \((q, 15 - q^2)\), \(h = 15 - q^2\) or \(y = 15 - q^2\) may be indicated on diagram	M1	3.1a
Completes rigorous argument to show the given result. It must be clear how they have defined the base and height with use of \(\dfrac{1}{2} \times 2q(15 - q^2)\) for whole triangle Or \(\left[\dfrac{1}{2}q(15 - q^2)\right] \times 2\) for two half triangles Or Explains why the area of the triangle is given by \(q(15 - q^2)\) with reference to the rectangle on either side of \(y\)-axis	R1	2.1
Deduces \(c = \sqrt{15}\) ACF	B1	2.2a
	(3)

Mark scheme (b)

Scheme	Marks	AO
Explains that maximum occurs when derivative equals 0 Condone incorrect variables in their derivative	E1	2.4
Differentiates w.r.t. \(q\) At least one term correct	M1	3.1a
Obtains \(15 - 3q^2\)	A1	1.1b
Solves ‘their \(\dfrac{\mathrm{d}A}{\mathrm{d}q}\)’ = 0 to find \(q\) and substitutes to find maximum area	M1	1.1a
Obtains correct maximum area ACF	A1	1.1b
Gives justification for maximum Could be evaluation of second derivative as \(-13.42\ldots \lt 0\) Or Test of gradient either side, Or Explanation, for example: This must be a max value as only turning point in the interval \(0 \lt q \lt \sqrt{15}\) and the area is 0 at the endpoints	E1	2.4
	(6)
	(9 marks)

Typical solution

\[\frac{\mathrm{d}A}{\mathrm{d}q} = 15 - 3q^2\]

Max occurs at \(\dfrac{\mathrm{d}A}{\mathrm{d}q} = 0\)

\[15 - 3q^2 = 0\]\[q = \sqrt{5}\]\[\frac{\mathrm{d}^2A}{\mathrm{d}q^2} = -6\sqrt{5} \lt 0 \text{ so local maximum}\]\[\therefore \text{Max area} = 15\sqrt{5} - 5\sqrt{5} = 10\sqrt{5}\]