If a closed cylindrical can is optimally designed (with respect to surface area used to hold a specified volume) can you determine how the diameter of the base is related to the height of the can?(please say as mush as you can and draw relevant pictures if you cant solve it) thaaanks youuu guyys!!
2007-11-26 03:10:39 · 4 answers · asked by blacksmith 3 in Science & Mathematics ➔ Mathematics
Πr^2h is the volume of a cylinder.
The relation btn the height(h) and diameter (D) can be given by the derivative with respect to the height or vice versa.
That is; v = Πr^2h ----------(i)
D = 2r thus r = D/2-----------(ii)
subst.(ii) into (i) then we have V = Π(D/2)^2h
Here we move onto get the derivative of V with respect to D
V = (ΠD^2h )/4
dV/dD = (2ΠDh )/4 = (ΠDh )/2---------(iii)
dV/dh = (ΠD^2 )/4 --------------------(iv)
Then do the multiplication of dv/dD * dh/dV //*Note that dh/dV is the inverse of dV/dh*//
dh/dV * dV/dD = 4/(ΠD^2) * (ΠDh)/2
This will give us 2h/D----------(v)
This I guess is what you were looking for my friend,I hope its the answer u had..
2007-11-26 03:40:59 · answer #1 · answered by Duse 3 · 0⤊ 0⤋
It's a maximization problem: for surface area A and volume V, maximize V/A.
If the cylinder has height h and radius r, A=2*pi*r*h + 2*pi*r^2 and V=2*pi*r^2*h
Hold A constant, rewrite h in terms of r, then maximize V in terms of r.
2007-11-26 03:34:49 · answer #2 · answered by Tom V 6 · 0⤊ 0⤋
It may be in your text book but a square is an optimized shape for maximum surface area. That is the only clue you will need to solve that problem.
2007-11-26 03:31:57 · answer #3 · answered by Brian 6 · 1⤊ 0⤋
Ask your teacher please. Calculus*
2007-11-26 03:15:23 · answer #4 · answered by Anonymous · 1⤊ 0⤋