# A cylinderical steel tank with an open top is to be constructed so that it will hold 512pi cubic meters of liquid.Find the radious and height of the tank that uses the least amount of material.

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To solve, apply the formula of volume of a cylinder.

`V=pir^2h`

Plug-in the given volume of the cylindrical tank.

`512pi=pir^2h`

Then, divide both sides by `pir^2` to isolate the h.

`(512pi)/(pir^2)=(pir^2h)/(pir^2)`

`512/r^2=h` (Let this be EQ1.)

Since the material of the cylindrical tank (with open top) must be minimized, set-up the equation of its area.

Area = Area of the bottom + Area of side

`A= pir^2+2pirh`

To express the right side as one variable, substitute EQ1.

`A=pir^2+2pir(512/r^2)`

`A=pir^2+(1024pi)/r`

To solve for r that would minimize the area, take the derivative of A.

`A=pir^2 + 1024pir^(-1)`

`A'=2pir-1024pir^(-2)`

`A'=2pir-(1024pi)/r^2`

Then, set A' equal to zero.

`0=2pir-(1024pi)/r^2`

To simplify the equation, divide both sides by 2pi.

`0/(2pi)=(2pir-(1024pi)/r^2)/(2pi)`

`0=r - 512/r^2`

And multiply both sides by r^2.

`r^2*0=(r - 512/r^2)*r^2`

`0=r^3-512`

Then, isolate r. So, add both sides by 512.

`0+512=r^3-512+512`

`512=r^3`

And take the cube root of both sides.

`root(3)(512)=root(3)(r^3)`

`8=r`

Next, solve for h. To do so, plug-in the value of r to EQ1.

`h=512/r^2=512/8^2=512/64=8` **Hence, to minimize the area of the cylindrical tank, its radius and height should be both 8 meters.**