A sheet of area `40m^2` is used to make an open tank with square base. Find the dimensions of the base such that the volume of this tank is maximum.

Explanation

Let r and h be the radius and height of the cylinder respectively.
Then the surface area (S) of the cylinder is given by
`S=2pir^(21)+2pirh`
`h=(S-2pir^(2))/(2pir)=(S)/(2pir)(1/r)-r`
Let v be the volume of the cylinder .Then
`V=pir^(2)h=pir^(2)[(S)/(2pi)(1/r)-r]=(Sr)/(23)-pir^(3)`
Then `(dv)/(dr)=(s)/(2)-3pir^(2),(d^(2)V)/(dr^(2))=-6pir`
Now `(dv)/(dr)=0 rarr (S)/(2)=3pir^(2)rarr r^(2)=(s)/(6pi)`
when `r^(2)=(S)/(6pi),"then" (d^(2)V)/(dr^(2))=-6pi(sqrt(s))/(6pi)lt0`
`therefore` By second derivative test the volume is the maximum when
`r^(2)=(S)/(6pi)`
Now when `r^(2)=(S)/(6pi),"then" h=(6pir^(2))/(2pi)(1/r)-r=3r-r=2r`
Hence the volume is the maximum when the height is two the radius i.e when the height is equal to the diameter.