Calculus Question?

Question

I'm having some difficulty solving this problem:

A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?

Here are some hints:
http://www.unr.nevada.edu/~saini/classes/math181F06/misc/ex4help.pdf

Answer 1

volume V = πR^2H + 2πR^3/3 (volume of cylinder plus volume of hemisphere)
area A = 2πRH + πR^2 + 2πR^2 = 2πRH + 3πR^2 (area around cylinder plus area of base plus surface area of hemisphere)

so from the first equation,
πR^2H = V - 2πR^3/3 or,
H = V/(πR^2) - 2R/3
substituting this into the area equation,
A = 2πR[ V/(πR^2) - 2R/3 ] + 3πR^2
A = 2V/R - 4πR^2/3 + 3πR^2 = 2V/R + 5πR^2/3
we want to minimize A, so differentiate this with respect to R and set = 0

A ' = -2V/R^2 + 10πR/3 = 0
10πR/3 = 2V/R^2
R^3 = 3V/(5π)
R = ³√[3V/(5π)]
and
H = V/(πR^2) - 2R/3 from above.

Answer 2

V = πr^2h + (4/6)πr^3
πr^2h = V - (2/3)πr^3
h = V/πr^2 - (2/3)r

A = πr^2 + 2πrh + 2πr^2
A = 2πrh + 3πr^2
A = πr(2h + 3r)
A = πr(2V/πr^2 - (4/3)r + 3r)
A = 2V/r + (5/3)πr^2
dA/dr = -2V/r^2 + π(10/3)r = 0
π(10/3)r = 2V/r^2
r^3 = 3V/5π
r = ³√(3V/5π)

h = V/πr^2 - (2/3)r

Answer 3

so the dimensions are to be minimized!

V of a cylinder = pi r^2 h+ 2/3 pi r^3

u can't solve this without any given or even relations of radius to height or anything!!!!!!!!! promise.
if u do have some given differentiate it. and equate to zero

Answer 4

S = πr² + 2πrh + 2πr²
= 3πr² + 2πrh
V = πr²h + ⅔πr³
So h = (V - ⅔πr³)/(πr²)
= (3V - 2πr³)/(3πr²)

So S = 3πr² + 2πr(3V - 2πr³)/(3πr²)
= 3πr² + 2(3V - 2πr³)/3r²
= 3πr² - (4/3)πr + 2V/r²

dS/dr = 6πr - (4/3)π - 4V/r³
= 0 for stationary points
So 6πr - (4/3)π - 4V/r³ = 0
ie 6πr^4 - (4/3)πr³ - 4V = 0

Let f(r) = 6πr^4 - (4/3)πr³ - 4V
f'(r) = 24πr³ - 4πr

By Newtons method

r n+1 = r n - f(r n)/f'(r n)

Try R1 = V/π
R2 = V/π - (6V^4/π³ - 4V³/π² - 4V)/(24V³/π² -4V²/π)
= V/π - [2V/π³ *(3V³ - 2πV² -2π³)]/[4V²/π² * (6V - π)
= V/π - [(3V³ - 2πV² -2π³)]/[2πV*(6V - π)]

Errrrrrrrrrrr this is getting real messy!!!!!!!!!
A particular value for V would have been nice!!!

Answer 5

ok V = h* area of bottom or h*(pi * r^2)

area top and bottom = 2 * (pi * r^2)
area of side is length of side which is circumference of circle * height

so area of side of tank is (2 * p1*r) * h
so whole area is 2 * (pi * r^2) + (2 * pi*r) * h
minimized differentiate
2*2* pi * r + 2*pi*h=0
=4pi*r+ 2 pi*h=0
2r = h
or d = h


i could be wrong but i bekieve top answer of other person looking at sphere not cylinder