SOLUTION: A company that manufactures dog food wishes to pack in closed cylindrical tins. What should be the dimensions of each tin if it is to have a volume of 128πcm³ and the minimum p

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Question 1207300: A company that manufactures dog food wishes to pack in closed cylindrical tins.
What should be the dimensions of each tin if it is to have a volume of 128πcm³
and the minimum possible surface area?
Found 4 solutions by ikleyn, greenestamps, Edwin McCravy, MathLover1:
Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
A company that manufactures dog food wishes to pack in closed cylindrical tin's as,
what should be the dimensions of each tin if it is to have a volume of 128π cm³
and the minimum possible surface area.
~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

As you know, the volume of a cylinder is 

V = pi%2Ar%5E2%2Ah, 

where pi = 3.14, r is the radius and h is the height.


In your case the volume is fixed:

pi%2Ar%5E2%2Ah = 128%2Api  cm^3.       (1)


The surface area of a cylinder is 

S = 2pi%2Ar%2Ah+%2B+2pi%2Ar%5E2,    (2)

and they ask you to find minimum of (2) under the restriction (1).


Using (1), I can rewrite (2) in the form

S(r) = %282pi%2Ar%5E2%2Ah%29%2Fr + 2pi%2Ar%5E2 = 2%2A%28128%2Api%29%2Fr%29 + 2pi%2Ar%5E2 = %28256%2Api%29%2Fr + 2pi%2Ar%5E2.   (3)


The plot below shows the function S(r) = 256%2Api%2Fr + 2pi%2Ar%5E2, and you can clearly see that it has the minimum.



    

        Plot y = %28256%2Api%29%2Fr+%2B+2%2Api%2Ar%5E2


To find the minimum, use Calculus: differentiate the function to get

S'(r) = -%28256%2Api%29%2Fr%5E2 + 4pi%2Ar = %28-256%2Api+%2B+4pi%2Ar%5E3%29%2Fr%5E2

and equate it to zero.


S'(r) = 0   leads you to equation  4%2Ar%5E3 = 256,   which gives 

r = root%283%2C64%29 = 4 cm.


Answer.  r = 4 cm, h = %28128%2Api%29%2F%28pi%2A4%5E2%29 = 8 cm gives the minimum of the surface area.

Solved.



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The volume of the cylindrical can with radius r and height h is to be 128pi cm^3:

V=%28pi%29r%5E2h=128pi [1]

The surface area of the can -- top, bottom, and side -- is

S=2%28pi%29r%5E2%2B2%28pi%29rh [2]

Solve [1] for h in terms of r and substitute in [2] to get an expression for the surface area in terms of the single variable r:

h=128%2Fr%5E2
S=2%28pi%29r%5E2%2B2%28pi%29r%28128%2Fr%5E2%29=2%28pi%29r%5E2%2B256%28pi%29%2Fr

Find the derivative of the expression for the surface area and set it equal to zero to find the radius r that minimizes the surface area:

dS%2Fdr=4%28pi%29r-256%28pi%29%2Fr%5E2
4%28pi%29r-256%28pi%29%2Fr%5E2=0
4%28pi%29%28r-64%2Fr%5E2%29=0
r-64%2Fr%5E2=0
r%5E3-64=0
r=4

The radius that minimizes the surface area is r=4; the corresponding height is 128/r^2 = 128/16 = 8.

ANSWER: radius 4cm, height 8cm

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
A company that manufactures dog food wishes to pack in closed cylindrical tins.
What should be the dimensions of each tin if it is to have a volume of
128πcm³ and the minimum possible surface area:

In the UK, you say "tin". In the US, we say "can". LOL 

The volume of a cylinder is 

V+=+pi%2Ar%5E2%2Ah+=+128%2Api 

Which simplifies to 
r%5E2%2Ah=128 
h=128%5E%22%22%2Fr%5E2
The surface area of a cylinder with a top and bottom is

A+=+2%2Api%2Ar%2Ah%2B2pi%2Ar%5E2
Substituting for h:
A+=+2%2Api%2Ar%28128%5E%22%22%2Fr%5E2%29%2B2%2Api%2Ar%5E2
A+=+%28256%2Api%29%2Fr+%2B+2%2Api%2Ar%5E2
A+=+256%2Api%2Ar%5E%28-1%29+%2B+2%2Api%2Ar%5E2
Differentiating w/r r:
dA%2Fdr=-256%2Api%2Ar%5E%28-2%29%2B4%2Api%2Ar
Setting the derivative equal to zero:
-256%2Api%2Ar%5E%28-2%29%2B4%2Api%2Ar=0
Dividing through by -4π
64r%5E%28-2%29-r=0
Multiplying through by r2
64-r%5E3=0
64=r%5E3
4=r

So the radius of each can should be 4 cm. (diameter = 8 cm.)
Substituting in
r%5E2%2Ah=128
4%5E2%2Ah=128
16%2Ah=128
h=8

And the height of each can should be 8 cm.

So the mid-cross section of each can should be an 8 cm x 8 cm square!





Edwin

Answer by MathLover1(20850) About Me  (Show Source):
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V=pi%2Ar%5E2%2Ah
surface area:
A=2pi%2Ar%2Ah%2B2pi%2Ar%5E2

given
V=128pi%2Acm%5E3
then
pi%2Ar%5E2%2Ah=128pi...simplify
r%5E2%2Ah=128
h=128%2Fr%5E2

substitute in formula for area
A=2pi%2Ar%28128%2Fr%5E2%29%2B2pi%2Ar%5E2
A=2pi%2A%28128%2Fr%29%2B2pi%2Ar%5E2
A=2pi%2A%28128%2Fr%29%2B2pi%2Ar%5E2
A=2pi%2Ar%5E2+%2B+%28256pi%29%2Fr

to minimize, derivate
first derivative
A′%28r%29=%284pi%2Ar%5E3+-+256pi%29%2Fr%5E2
A′%28r%29=4pi%2Ar-256pi

second derivative
A′'%28r%29=4pi
now, A′′%284%29%3E0 at r+=+4 is point of minima

substitute r=4+in h=128%2Fr%5E2
h=128%2F%284%29%5E2
h=128%2F16
h=8
so, minimum surface area the can should have a radius of 4cm+and height of 8cm