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posted by  chibba on 1/22/2008 11:05:56 PM  |  status: Live  

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Course Textbook Chapter Problem
Calculus N/A N/A N/A
Question Details:
An oil can is to be made in the form of a right circular cylinder to contain 16π in3. What dimensions of the can will require the least amount of material?
Do not use L'Hopital's Rule on my post unless it's neccessary!
 
Feel free to msg me if you have any questions with my responses!
 
The indeterminate forms include 00, 0/0, 1, ∞ - ∞, ∞/∞, 0×∞, and ∞0.

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posted by Yashar Bahman on 1/22/2008 11:40:52 PM  |  status: Live
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chibba's comment:
"ty"
Response Details:
ahh i was just doing this last week.

okay so this problem requires optimization.

the first thing we want to do is find the formula for the volume and surface area of a cylinder.
v = πr2h
sa = 2πr2 + 2πrh

now we know the volume must equate to 16π (given) therefore 16π = πr2h

simplified that becomes:

cross out the pi's:

now plug that into the original area formula: 

sa = 2πr2 + 2πr(16/r2)

then simplify it to: 2πr(r2+16/r)

now we must derive! 2πr(2r-16/r2) (using the chain rule for r2 and quotient rule for 16/r)

set it equal to zero and divide out 2πr: 0 = 2r-16/r2

simplify more: so r = 8/r2

multiply both sides by r2

so r3 = 8 therefore r = 2

plug that back into
and find that h = 4

the answer is r = 2 inches and h = 4 inches
Tags: Math, Calculus
Sage
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posted by Bart_Boy on 1/22/2008 11:50:29 PM  |  status: Live
Asker's Rating: Helpful   
chibba's comment:
"ty"
Response Details:


solve the Volume equation for h

Substitute into the surface area equation

take the derivative with respect to r

The only real zero of this derivative is r = 2
here the sign of the derivative switches from negative to positive, so it is a minimum

back to the equation solved for h, and we get
r = 2 in
h = 4 in

Tags: Math, Calculus
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