Question
Asked Oct 24, 2019
Suppose a cylindrical can is being made to hold 1 liter of oil. Design the can so
that the minimum amount of metal can be used. The area of the metal is given by
1000
A=27r27 rh Note that 1L = 1000cc so zr h
1000, h
2
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Suppose a cylindrical can is being made to hold 1 liter of oil. Design the can so that the minimum amount of metal can be used. The area of the metal is given by 1000 A=27r27 rh Note that 1L = 1000cc so zr h 1000, h 2

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check_circleExpert Solution
Step 1

Area of the metal is fixed. And h=1000/(pi*r^2)

We plug h in area and simplify and express A in terms of r.

А-2лг ?+2лrh
1000
А32лт "+2лг
лг
2000
?+
А-2лг
г
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А-2лг ?+2лrh 1000 А32лт "+2лг лг 2000 ?+ А-2лг г

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Step 2

For minimum metal dA/dr=0. Find dA/dr.

2000
2
A-27r
r
2000
dA
d
2лт +
dr
dr
r
dA
2000
- 2n(2r)
dr
r2
dA
=4nr
dr
2000
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2000 2 A-27r r 2000 dA d 2лт + dr dr r dA 2000 - 2n(2r) dr r2 dA =4nr dr 2000

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Step 3

 Set dA/dr equal to 0. From ther...

2000
0=47r
r2
2000
4Tr
2
r
2000
4T
r3159.1549431
r=159.1549431
r5.41926
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2000 0=47r r2 2000 4Tr 2 r 2000 4T r3159.1549431 r=159.1549431 r5.41926

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Math

Calculus