O lvl A Maths: Applications of Differentiation (Maxima and Minima)

RI 1999 Sec 4 Common Test Q11

ABCD is a trapezium where AB is parallel to DC, AB = 10 cm, AD = BC = 8 cm and ∡ADC ∡BCD = θ radians. Prove that the area of the trapezium, S cm² is given by S = 16 sin θ (4 cos θ + 5).

Hence, calculate the value of θ for which S has a stationary value. Determine whether this value of S is a maximum or minimum. (12 marks)

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Answer:

Area of trapezium
= area of the 2 side triangles + area of centre rectangle
= 2 * (½ * 8 cos θ * 8 sin θ) + 10 * 8 sin θ
= 64 sin θ cos θ + 80 sin θ
= 16 sin θ (4 cos θ + 5)
(shown)

Differentiate S w.r.t. θ
$\frac{dS}{d\theta }=16\sin \theta \left(-4\sin \theta \right)+\left(4\cos \theta +5 \right)16\cos \theta$
$\frac{dS}{d\theta }=-64\sin^{2} \theta+64\cos^{2} \theta+80\cos \theta$
$\frac{dS}{d\theta }=-64(1-\cos^{2} \theta)+64\cos^{2} \theta+80\cos \theta$
$\frac{dS}{d\theta }=-64+64\cos^{2} \theta+64\cos^{2} \theta+80\cos \theta$
$\frac{dS}{d\theta }=128\cos^{2} \theta+80\cos \theta -64$

Set dS/dθ = 0 to find stationary value
$128\cos^{2} \theta+80\cos \theta -64=0$
$8\cos^{2} \theta+5\cos \theta -4=0$
cos θ = 0.4605823 or -1.0855823 (reject)
θ = 62.6°

$\frac{d^{2}S}{d\theta ^{2}}=256\cos \theta \left(-\sin \theta\right)-5\sin \theta$
When θ = 62.6°,
$\frac{d^{2}S}{d\theta ^{2}}=-109<0$

Hence, the value of S is a maximum.

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O lvl A Maths: Applications of Differentiation (Maxima and Minima)