SOLUTION: A company needs to make a cylindrical can that can hold precisely 0.7 liters of liquid. If the entire can is to be made out of the same material, find the dimensions of the can tha

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Question 1001753: A company needs to make a cylindrical can that can hold precisely 0.7 liters of liquid. If the entire can is to be made out of the same material, find the dimensions of the can that will minimize the cost.
Found 2 solutions by ankor@dixie-net.com, KMST:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A company needs to make a cylindrical can that can hold precisely 0.7 liters of liquid.
If the entire can is to be made out of the same material, find the dimensions of the can that will minimize the cost.
:
.7 liters = .7*1000 700 cubic/cm
let r = the radius of the can that has a volume of 700 cm/cm
:
pi%2Ar%5E2%2Ah = 700
h = 700%2F%28%28pi%2Ar%5E2%29%29
h = 222.817%2Fr%5E2
The surface area
S.A. = 2%28pi%2Ar%5E2%29 + 2%2Api%2Ar%2Ah
Factor out 2*pi*r
S.A. = 2pi%2Ar%28r+%2B+h%29
replace h with 222.817/r^2
S.A. = 2pi%2Ar%28r+%2B+%28222.817%2Fr%5E2%29%29
:
Graph this equation, radius on the x axis, Surface area on the y axis
+graph%28+300%2C+200%2C+-4%2C+10%2C+-200%2C+1000%2C+x-2%2C+6.28x%5E2%2B%281400%2Fx%29%29+
:
Radius for minimum surface area about 4.8 cm
Find the height
h = 222.817/4.8^2
h = 9.67 cm
:
Check: find the vol with these values
V = pi%2A4.8%5E2%2A9.67
V = 700 cu/cm

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The volume of the can must be
V=0.7L=700mL=700cm%5E3
Let's define the measurements
R= radius of the base of the can, in cm
h= height of the can, in cm
As functions of those two variables, the Volume (in cubic centimeters) and the total surface Area of the can (in square centimeters) are
V=pi%2AR%5E2%2Ah and A=2pi%2AR%5E2%2B2pi%2AR%2Ah .
Since the volume of the can must be 700cm%5E3 ,
system%28V=700%2CV=pi%2AR%5E2%2Ah%29--->pi%2AR%5E2%2Ah=700--->h=700%2F%28pi%2AR%5E2%29
Substituting the expression fond for h in A=2pi%2AR%5E2%2B2pi%2AR%2Ah , we get
A=2pi%2AR%5E2%2B2pi%2AR%2A%28700%2F%28pi%2AR%5E2%29%29
A=2pi%2AR%5E2%2B1400%2FR

THE CALCULUS SOLUTION:
The minimum for A=2pi%2AR%5E2%2B1400%2FR occurs for a value of R that makes the derivative dA%2FDR zero.
dA%2FdR=4pi%2AR%2B1400%2A%28-1%2FR%5E2%29
dA%2FdR=%284pi%2AR%5E3-1400%29%2FR%5E2%29
dA%2FdR=0--->4pi%2AR%5E3-1400=0--->R%5E3=1400%2F4pi--->R%5E3=350%2Fpi--->R=root%283%2C350%2Fpi%29 or R=%28350%2Fpi%29%5E%28%221+%2F+3%22%29
For R%3Croot%283%2C350%2Fpi%29 , dA%2FdR%3C0 and A is decreasing.
For R%3Eroot%283%2C350%2Fpi%29 , dA%2FdR%3E0 and A is increasing.
For R=root%283%2C350%2Fpi%29 , dA%2FdR=0 and A is minimum.
So, the dimensions of the can that will minimize the cost are
highlight%28R=root%283%2C350%2Fpi%29%29highlight%28cm=about4.8cm%29 and
h=700%2F%28pi%2A%28350%2Fpi%29%5E%28%222+%2F+3%22%29%29--->highlight%28h=2%2A%28350%2Fpi%29%5E%281%2F3%29%29highlight%28cm=about9.6cm%29 .