The dimensions of the cylinder can be used to minimise cost of manufacture.
Let's ignore the metal's thickness and assume that the material cost to manufacture is precisely proportionate to the surface area of a perfect cylinder.
A=2πr(r+h)
Given that V=1000=r2h and h=1000=r2, we can write
A=2πr(r+1000πr2)
A=2πr2+2000r−1
By setting the derivative to zero, we may determine the value of r that minimises A:
A′=4πr−2000r−2
0=4πr−2000r−2
2000r−2=4πr
2000=4πr3
r=500π−−−√3
r=5.419 + cm
h=1000πr2=10.838 + cm
The can with the smallest surface area has a volume of 1000 cm 3 and measures 5.419+ cm in radius and 10.838+ cm in height. The can has a surface area of 553.58 cm 2. Given a constant volume, the cylinder with diameter equal to height has the least surface area.
Can surface area (cm2) versus. radius (cm), where capacity = 1000cm 3.
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