A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the costs of manufacture the can. Enter the value of r.
For this problem the correct answer is 5.41926, but I keep getting 0.541926.
Try to do it yourself, and see if you get the same as I did or the correct answer. Please show the procedure in an organized and easy-to-understand way. If you got the same answer as I did, I'd appreciate you let me know. No procedure is necessary if you got the same answer as I did.
For this problem the correct answer is 5.41926, but I keep getting 0.541926.
Try to do it yourself, and see if you get the same as I did or the correct answer. Please show the procedure in an organized and easy-to-understand way. If you got the same answer as I did, I'd appreciate you let me know. No procedure is necessary if you got the same answer as I did.
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1 liter = 1000 cubic centimeters; since we'll want to be able to measure our dimensions in centimeters, use V = 1000
V = pi r^2 h = 1000
h = 1000/(pi r^2)
Surface area A = the area of the 2 circular bases plus the lateral area = 2 pi r^2 + 2 r pi * h
A = 2 pi r^2 + 2 r pi (1000/pi r^2)
A = 2 pi r^2 + 2000/r
A' = 4 pi r - 2000/r^2 = 0
4pi r = 2000/r^2
4 pi r^3 = 2000
r^3 = 500/pi
r = 5.419 cm
V = pi r^2 h = 1000
h = 1000/(pi r^2)
Surface area A = the area of the 2 circular bases plus the lateral area = 2 pi r^2 + 2 r pi * h
A = 2 pi r^2 + 2 r pi (1000/pi r^2)
A = 2 pi r^2 + 2000/r
A' = 4 pi r - 2000/r^2 = 0
4pi r = 2000/r^2
4 pi r^3 = 2000
r^3 = 500/pi
r = 5.419 cm
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1 liter = 1000 cubic cm
If V= can volume
A= can surface area
h= can height
r= radius of circular can end
Then
V= pi x r^2 x h
A= 2 x (pi x r^2) + (2 x pi x r x h) , area of the two ends + area of can side
Either by making a table for various values of r and h and the A that results or
by using calculus to find the value of r & h which minimizes A ( set first derivative of A with respect to r equal to 0 and solve :
note that h is eliminated by using the equation for V so that h = 1000 / [pi x r^2] )
It becomes obvious that A is minimized when the diameter of the end is the same as the height, that is when h=2 x r
So substituting that into the volume equation :
V= pi x r^2 x (2 x r)
V= 2 x pi x r^3
solving for r
r = CubeRoot of V divided by two pi = { V / ( 2 x pi ) } ^(1/3)
After substituting 1000 for V you will get r = 5.41926 cm
So the can has dimensions of r = 5.41926 cm, d=h= 10.83852 cm
If V= can volume
A= can surface area
h= can height
r= radius of circular can end
Then
V= pi x r^2 x h
A= 2 x (pi x r^2) + (2 x pi x r x h) , area of the two ends + area of can side
Either by making a table for various values of r and h and the A that results or
by using calculus to find the value of r & h which minimizes A ( set first derivative of A with respect to r equal to 0 and solve :
note that h is eliminated by using the equation for V so that h = 1000 / [pi x r^2] )
It becomes obvious that A is minimized when the diameter of the end is the same as the height, that is when h=2 x r
So substituting that into the volume equation :
V= pi x r^2 x (2 x r)
V= 2 x pi x r^3
solving for r
r = CubeRoot of V divided by two pi = { V / ( 2 x pi ) } ^(1/3)
After substituting 1000 for V you will get r = 5.41926 cm
So the can has dimensions of r = 5.41926 cm, d=h= 10.83852 cm
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V = pi r^2 h = 1000 cm^3
=> h =1000 / (pi r^2)
A = 2pi r^2 + 2pi r h
=> A = 2pi r^2 + 2pi r [1000 / (pi r^2)]
=> A(r) = 2pi r^2 + 2 [1000 / r]
=> dA/dr = 2[ 2pi r - 1000 / r^2] = 0
=> 2pi r = 1000 / r^2
=> r^3 = 1000/ 2pi = 500 / pi
=> r = 5.41926 cm
I don't know where you've done a mistake , but that's the correct answer .
=> h =1000 / (pi r^2)
A = 2pi r^2 + 2pi r h
=> A = 2pi r^2 + 2pi r [1000 / (pi r^2)]
=> A(r) = 2pi r^2 + 2 [1000 / r]
=> dA/dr = 2[ 2pi r - 1000 / r^2] = 0
=> 2pi r = 1000 / r^2
=> r^3 = 1000/ 2pi = 500 / pi
=> r = 5.41926 cm
I don't know where you've done a mistake , but that's the correct answer .
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not sure