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

Answer 1

Ignoring the fact that a cuboid is not a good shape as a can:

As the surface area of the cuboid can is fixed, the side length of the square base and height of the can are in an inverse relationship: as one increases the other decreases.

Neither can go negative.

When the side length 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 side length 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 side length, which allows the maximum to be found (when the volume differentiated in terms of height or side length (that is dV/dh or dV/ds) 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.

----------------------------------------------------------------------------------------------------------------

Surface area (A) = 2s² + 4sh

→ h = (A - 2s²)/4s

Volume (V) = s²h

= s²(A - 2s²)/4s

= As/4 - s³/2

dV/ds = 0 when V is a maximum

dv/ds = d/ds (As/4 - s³/2)

= A/4 - 3s²/2

→ s = (A/6)^(1/2)

v = As/4 - s³/2

→ Vmax = A( (A/6)^(1/2) )/4 - (A/6)^(3/2)/2

= 6(A/6)^(3/2)/4 - (A/6)^(3/2)/2

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

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

Vmax = (450cm²/6)^(3/2)

= (75)^(3/2) cm³

≈ 649.5 cm³