A cylindrical can has prescribed surface area S. What dimensions for the can yield the maximum volume?
November 22nd, 2009 | by admin |please show me how to work this problem using tools you learn from Calculus 3
Let r be the radius of the base circle & h the height of
the cylindrical can. Then
2pir^2+2pirh=S=> (pi=3.141592…, the can-cover is counted)
h=[S/(2pir)-r]—–(1)
The volume V of the cylindrical can is given by
V=pir^2h=>
V=pir^2[S/(2pir)-r] (from (1))=>
V=rS/2-pir^3=>
V’=S/2-3pir^2=>–(2)
V"=-6pir———–(3)
V’=0=>
r=sqrt(S/(6pi))——(4)
from (3),(4),get
V"(S/(6pi))<0=> (4) gives a max.V
From (1) & (4), get
h=sqrt(6piS)/(3pi)
2 Responses to “A cylindrical can has prescribed surface area S. What dimensions for the can yield the maximum volume?”
By PINKGREEN on Nov 22, 2009 | Reply
Let r be the radius of the base circle & h the height of
the cylindrical can. Then
2pir^2+2pirh=S=> (pi=3.141592…, the can-cover is counted)
h=[S/(2pir)-r]—–(1)
The volume V of the cylindrical can is given by
V=pir^2h=>
V=pir^2[S/(2pir)-r] (from (1))=>
V=rS/2-pir^3=>
V’=S/2-3pir^2=>–(2)
V"=-6pir———–(3)
V’=0=>
r=sqrt(S/(6pi))——(4)
from (3),(4),get
V"(S/(6pi))<0=> (4) gives a max.V
From (1) & (4), get
h=sqrt(6piS)/(3pi)
References :
By gile on Nov 22, 2009 | Reply
The total surface area of the cylindrical can:
S = 2πRh + 2πR² = 2πR(R + h)
The height of the can in function of R
h = (S/2πR) - R
The volume of the can
V = πR² h = (SR/2) - πR³
S being a constant,
dV/dR = S/2 - 3πR²
d²V/dR² = -3πR
d²V/dR² < 0 for all R > 0, which is the case of a length.
Thus V admits a maximum when dV/dR = 0.
dV/dR = 0 ==> S/2 - 3πR² = 0 or
S = 6πR²
Together with the relation established above
S = 2πR(R + h),
we get
2πR(R + h) = 6πR²
R + h = 3R
h = 2R
Conclusion: For a prescribed (fixed) surface S, the can yields the maximum volume if its height (h) is equal to its diameter (2R)
References :