A cylindrical can that has a capacity of 20 m3 will be made. The metal used to build the top costs $10 per square meter while the bottom costs $20 per square meter. The material used for the side costs $15 per square meter. What are the dimensions that will minimize the cost?.

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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