it is required to manufacture a cylinderical water tank with both ends closed. the height should be 0.8m. a steel sheet of area = 4m^2 (excluding all fabrication allowances) is avaiable for the purpose. calculate the diameter of the cylinder if the thickness of the tank can be neglected.
2007-05-03 02:29:18 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Total surface area of cylinder = 2pi.r^2 + 2pi.r.h = 4
h=0.8
2pi.r^2 + 1.6pi.r - 4 = 0
r= -1.6pi±sqrt[(1.6pi)^2 + 4*2pi*4]/(4pi)
=-1.6pi±11.21/(4pi)
=0.49 (approx) ------ taking positive root as you cannot have a negative radius
diameter = 0.98m or about 98cm
2007-05-03 02:41:29 · answer #1 · answered by gudspeling 7 · 1⤊ 0⤋
Surface of 1 base=pixRxR
Surface of cylinder area = base perimeter x 0.8
= 2xpixRx0.8
= 1.6xpixR
Total surface = 2xbase + cylinder
= 2xpixRxR+1.6xpixR
You must solve the equation 2xpixRxR+1.6xpixR=4
Equation:
2xpixRxR+1.6xpixR-4=0
Delta=(1.6xpi)x(1.6xpi)+32pi
Delta~105.5575... (there are 2 solutions!)
2 solutions to this equation:
R1=-1.217... (invalid in your case)
R2=0.4175...
You can manufacture a cylinder of Radius 0.4175m
Therefore Diameter is 0.8351m
(results have been rounded to the 4th decimal - mm -)
2007-05-03 03:18:49 · answer #2 · answered by Rosamonte2000 2 · 0⤊ 1⤋
Let s = surface area
Let r = radius
Let h = height = 0 . 8 m
s = 2 π r² + 2 π r h
4 = 6.28 r² + 5.024 r
6.28 r² + 5.024 r - 4 = 0
r = [- 5.024 ± √(25.2 + 100.5)] / 2
r = [- 5.024 ± 11.2] / 2
r = 0.49 m (taking +ve value for r)
Diameter = 0.98 m = 98 cm
2007-05-03 19:51:03 · answer #3 · answered by Como 7 · 0⤊ 0⤋
thats so easy...just plug in the given numbers into the equation for a diameter, then divide by 2 or formuls for a radius...
YOU NEED AN AVATAR.
2007-05-03 02:33:27 · answer #4 · answered by Anonymous · 0⤊ 1⤋