SOLUTION: Hi Everyone, stuck in again on a weekend doing my maths !, having problems with this one please can someone help A circular-cylindrical oil drum is required to have a given surf

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Hi Everyone, stuck in again on a weekend doing my maths !, having problems with this one please can someone help A circular-cylindrical oil drum is required to have a given surf      Log On


   

Question 376236: Hi Everyone, stuck in again on a weekend doing my maths !, having problems with this one please can someone help
A circular-cylindrical oil drum is required to have a given surface area S (including its lid and base), ie S is a constant. Find the proportions of the design which contain the greatest volume V .
Thank You
John
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A circular-cylindrical oil drum is required to have a given surface area S
(including its lid and base), ie S is a constant.
Find the proportions of the design which contain the greatest volume V .
:
Surface area: S = 2(pi*r^2) + (2*pi*h)
:
Volume: V = pi*r^2*h
:
Find the relationship between the surface area and the volume
:
V%2FS = %28pi%2Ar%5E2%2Ah%29%2F%28%282pi%2Ar%2Ah%29%2B+%282%2Api%2Ar%5E2%29%29 = %28pi%2Ar%5E2%2Ah%29%2F%282%2Api%2Ar%28h%2Br%29%29
Cancel pi*r
V%2FS =%28r%2Ah%29%2F%282%2A%28h%2Br%29%29
Therefore
S = 2*(h+r)
Assume a value: S = 120
2h + 2r = 120
h + r = 60
h = (60-r)
:
V = pi*r^2*(60-r): replace h with (60-r)
Find Max volume. graph the equation y = 3.14*x^2*(60-x)
+graph%28+300%2C+200%2C+-20%2C+100%2C+-20000%2C+110000%2C+3.14%2Ax%5E2%2A%2860-x%29%29+
Max volume occurs when r = 40
then h = 60 - 40
h = 20
:
The proportion:
r%2Fh = 40%2F20,
we can say max volume when radius:height = 2:1