Application of Maximisation and Minimisation (1 Viewer)

[z600]

Humm i came across this question in my homework. I thought it was relatively easy but i just cant get it...

Show that a closed cylindrical can of fixed volume will have minimum surface area when its height is equal in length to its diameter.

thankss

 

[undalay]

Heres my proof: There might be a cleaner way, but this is my method.

SA = 2 pi R^2 + 2 pi R h
V = pi R^2 h
Now since volume is not a specific case, you can set it to anything. I chose 100 for convenience.

100 = pi R^2 h
R^2 = 100 / hpi

Subin SA

SA = 2pi(100/hpi) + 2pih(root(100/hpi))
Simplifies to:
SA = 200/h + 20root(pih)
SA' = -200/h^2 + 10pi^1/2h^-1/2 = O
Times everything by h^2
0 = -200 + 10pi^1/2 h ^1/1/2
20/root(pi) = h^1/1/2
400/pi = h^3
h = cuberoot(400)/cuberoot(pi)
h = 5.0308

~insert minimum turning point proof ~ Just sub into SA'

Now subin to volume formula
r = 10/root(pih)
2r = 20/root(pih)
2r = 5.0308

h = 2r

 

[webby234]

V = piD²h/4
A = piD²/2 + piDh

h = 4V/piD²

A = piD²/2 + 4VpiD/piD²
A = piD²/2 + 4V/D
dA/dD = piD - 4V/D² (V is constant)
d²A/dD² = pi + 8V/D³ which is always positive so any stationary points are minimums.

dA/dD = piD - 4V/D² = 0
piD = 4(piD²h/4)/D²
piD = pih
D = h

 

[z600]

V = piD²h/4
A = piD²/2 + piDh

h = 4V/piD²

A = piD²/2 + 4VpiD/piD²
A = piD²/2 + 4V/D
dA/dD = piD - 4V/D² (V is constant)
d²A/dD² = pi + 8V/D³ which is always positive so any stationary points are minimums.

dA/dD = piD - 4V/D² = 0
piD = 4(piD²h/4)/D²
piD = pih
D = h

very nicee thanks