Question

A cylindrical container of height equal to twice the diameter of its base can

hold 12 liters (1L= 1,000 cm3) of water. Another cylindrical container with

the same capacity has its height equal to three times the diameter of its

base. Where the diameter of 1st container is 19.69cm and the diameter of 2nd container is 22.55cm

1. Determine the amount of aluminum required for making the first

container?

2. Determine the amount of aluminum required for making the

second container? 

Answer 1

Assuming the containers are closed;

Total surface area (TSA) =$2\pi rh+2\pi r^2$

Where;

r=radius=half the diameter

h=height

1.amount of aluminum = $2\pi rh+2\pi r^2$

In this case, r=19.69cm/2=9.845cm

h=19.69cm*2=39.38cm

Therefore ;

TSA=$2*\pi *9.845cm*39.38cm+2*\pi *(9.845cm)^2=3044.958cm^2$

2.here,r=22.55cm/2=11.275cm

h=22.55cm*3=67.65cm

Therefore ;

TSA=$2\pi rh+2\pi r^2$

$=2*\pi *11.275cm*67.65cm+2*\pi *(11.275cm)^2=5591.277cm^2$