How do you find the largest possible capacity of a closed cylindrical can with surface area 450cm2?

Answer 1

As the surface area of the cylindrical can is fixed, the radius and height of the can are in an inverse relationship: as one increases the other decreases.

Neither can go negative.

When the radius is 0, the volume is zero

When the height is 0, the volume is zero

Somewhere in between the volume reaches a maximum value.

To solve this, you need to specify the radius in terms of the height (or vice-versa), which can be done form the fixed surface area of the can.

The volume can then be specified in terms of only either the height or radius, which allows the maximum to be found (when the volume differentiated in terms of height or radius (that is dV/dh or dV/dr) is equal to zero).

You now have the How-to to solving the problem. Have a go before looking at applying the How-to requested (above) to the given surface area below.

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Surface area (A) = 2πr² + 2πrh

→ h = (A - 2πr²)/2πr

Volume (V) = πr²h

= πr²(A - 2πr²)/2πr

= r(A - 2πr²)/2

= Ar/2 - πr³

dV/dr = 0 when V is a maximum

dV/dr = d/dr (Ar/2 - πr³)

= A/2 - 3πr²

→ r = (A/(6π))^(1/2)

V = Ar/2 - πr³

→ Vmax = (A/2)( (A/(6π))^(1/2) ) - π(A/(6π))^(3/2)

= (6π/2)( (A/(6π))^(3/2) ) - π(A/(6π))^(3/2)

= 2π(A/(6π))^(3/2)

The value of A at 450 cm² can now be used to find the maximum volume:

Vmax = 2π(450cm²/(6π))^(3/2)

= 2π(75/π))^(3/2) cm³

≈ 1837 cm³