# a cylindrical steel tank with an open top is to be constructed so that it will hold 512pi cubic meters of liquid.a cylindrical steel tank with an open top is to be constructed so that it will hold...

a cylindrical steel tank with an open top is to be constructed so that it will hold 512pi cubic meters of liquid.

a cylindrical steel tank with an open top is to be constructed so that it will hold 512pi cubic meters of liguid. Find the radius and height of the tank that uses the least amount of material.

Note: a) find the equation to maximized or minimized.

b) finding the solution.

c) showing that your solution is an absolute maximum or minimum.

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### 1 Answer

Let the radius of the cylinder is r and height is h.

Volume of cylinder `= 512m^3`

`pi*r^2*h = 512`

` h = 512/(pi*r^2)`

The amount of material needed can be represent by the area of the material.

Surface area of cylinder `(A) = pi*r^2+2*pi*r*h`

`A = pi*r^2+2*pi*r`

`A = pi*r^2+2*pi*r*512/(pi*r^2)`

`A = pi*r^2+1024/r`

When the material amount is least or maximum then `(dA)/(dr) = 0`

`(dA)/dr = 2pi*r+1024*(-1/r^2)`

When `(dA)/(dr) = 0;`

`2pi*r+1024*(-1/r^2) = 0`

`r^3 = 512/pi`

`r = 5.462`

When A is a minimum then `(d^2A)/(dA^2) >0` at` r = 5.462`

`(d^2A)/(dA^2)`

`= 2pi+1024*2/r^3`

since r>0; `(d^2A)/(dA^2)>0 ` always. So we have a minimum for A at `r = 5.462`

*Radius of cylinder for minimum material = 5.462*

**Height of cylinder for minimum material = `512/(pi*5.462^2)` = 5.462**

*So the height and radius of the cylinder is equal when the area of the material is minimum.*

Note:

It is assumed that the area of material for joints and welds are negligible or constant in all cases.

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