The dimensions of the cylinder can be used to minimise cost of manufacture.
<h3>How to we find minimised cost?</h3>
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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