Solution of Past Paper of CAIE Math 9709/13 Oct/Nov 2019 Question # 5

CIE/GCE/AS/Math/P1/19/Nov/13/Q#5

Question




The dimensions of a cuboid are x cm, 2x cm and 4x cm, as shown in the diagram.

i. Show that the surface area S cm 2 and the volume V cm3 are connected by
the relation
2
𝑆 = 7𝑉3

ii. When the volume of the cuboid is 1000 cm 3 the surface area is increasing
at 2 cm2 s−1. Find the rate of increase of the volume at this instant.


Solution
i.


We are given dimensions of the cuboid as dimensions of a cuboid are x cm, 2x cm and
4x cm.


We are required to find a relation between surface area 𝑆 and volume of 𝑉.


First, we find surface area 𝑆 of cuboid.


Expression for the surface area of the box is;

, CIE/GCE/AS/Math/P1/19/Nov/13/Q#5

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐵𝑜𝑥 = 2 × (𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ + 𝑙𝑒𝑛𝑔𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡 + ℎ𝑒𝑖𝑔ℎ𝑡 × 𝑤𝑖𝑑𝑡ℎ)


𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐵𝑜𝑥 = 2 × (𝑙𝑤 + 𝑙ℎ + ℎ𝑤 )


Therefore, surface area 𝑆 of the given cuboid is;


𝑆 = 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑢𝑏𝑜𝑖𝑑 = 2 × (𝑙𝑤 + 𝑙ℎ + ℎ𝑤 )


𝑆 = 2 × ((4𝑥 )(2𝑥) + (4𝑥 )(𝑥 ) + (𝑥 )(2𝑥))


𝑆 = 2 × (8𝑥2 + 4𝑥 2 + 2𝑥 2 )


𝑆 = 2 × (14𝑥2 )


𝑆 = 28𝑥2


Next, we find volume 𝑉 of cuboid.


Expression for the volume of the rectangular block is;


𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝐵𝑙𝑜𝑐𝑘 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑙𝑤ℎ


Therefore, volume 𝑉 of the given cuboid is;


𝑉 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝐵𝑙𝑜𝑐𝑘 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑙𝑤ℎ


𝑉 = (4𝑥 )(2𝑥 )(𝑥 )