Search 75,752 tutors
FIND TUTORS
Ask a question
0 0

Optimization Problems

Tutors, please sign in to answer this question.

2 Answers

A cylindrical can is to have a volume of 1L.
a) Find the height and radius of the can that will minimize the surface area
 
1 liter = 1000 cm3
 
Let x = the radius of the top and bottom of the can
Let y = the height of the can
 
The volume of the can is:  V = (pi)x2y = 1000 cm3.  So y = 1000/(pi)x2
 
The surface area of the can is: S = 2(pi)x2 + 2(pi)xy
 
Substituting 1000/(pi)x2 for y, the surface area is:
 
S = 2(pi)x2 + 2(pi)x*1000/(pi)x2 = 2(pi)x2 + 2000x-1
 
Take the derivative of S wrt x:
 
S' = dS/dx = 4(pi)x - 2000x-2
 
S' = 0 = 4(pi)x - 2000x-2
 
2000x-2 = 4(pi)x
 
2000/4(pi) = x3
 
x = (2000/4(pi))1/3 = 10(1/2(pi))1/3 ≈ 5.42 cm
 
y = 1000/(pi)x2 = 10.84 cm
 
Check:
 
V = (pi)(5.42)2(10.84) = 1000
 
b) What is the ratio of the height to the diameter?
 
The ratio of the height (y) to the diameter (2x) = 1
 
Do pop cans have a similar ratio? If not, why?  You're on your own for this question!  ;-)
This is a cylinder so we know that the base and top are A = π*r2 where r is the radius of the cylinder. The side area is the length of the outer edge of the base times the height (which we will call h) or A = 2πr*h. So the total area is
 
A = 2π*r2 + 2πr*h = 2π (r2 + rh).
 
The volume is just:
 
V = πhr2 = 1
 
So, using the volume we have h = 1/(πr2). We can substitute this into Area to simplify our expression.
 
A = 2π (r2 + 1/(πr)).
 
Now we take the derivative of A with respect to r, so that we can find critical points:
 
A' = 2π ( 2r - 1 / (πr2) ).
 
Set it equal to zero to find critical points, discarding the constant 2π since it is irrelevant:
 
0 = 2r - 1 / (πr2),
 
2r = 1 / (πr2),
 
r3 = 1/(2π),
 
r = 1 / (2π)1/3.
 
So our diameter is
 
D = 2πr = 2π * 1 / (2π)1/3 = (2π)2/3.
 
Now put this in the volume equation to find h.
 
πh(1/ (2π)1/3)2 = 1,
 
πh/ (2π)2/3 = 1,
 
h= 22/31/3.
 
Now we can find the ratio:
 
Ratio = h/D = (22/31/3 ) / (2π)2/3 = ( 1/π1/3 ) / π2/3 = 1 / π
 
 

Comments

Sorry for the colliding answer - we posted almost at the same time.  ;-)
As long as we're all having fun!

Comment