You are designing a 1000 cm^{3} right circular cylindrical can w

William Cleghorn 2021-12-18 Answered
You are designing a 1000cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be
A=8r2+2πrh
What is the ratio now of h to r for the most economical can?
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kaluitagf
Answered 2021-12-19 Author has 38 answers
Step 1
The volume of a right circular cylinder is V=πr2h where r is the radius of the circular base of the cylinder and h is the height of the cylinder. Since we have to take waste into account, the amount of aluminum used up by the can A=8r2+2πrh. In order to find the minimum amounts of aluminum for the can hold a volume of 100cm3, we must express A in terms of one variable only. Let's express A in terms of r. So have to express h in terms of r and we know that πr2h=1000. Thus we have:
πr2h=1000
h=1000πr2
So the surface area in terms of r is A=8r2+2πr(1000πr2)=8r2+2000r
Now let's differentiate A with respect to r and set it to 0 and solve for r:
dAdr=16r2000r2
0=16r2000r2
2000r2=16r
16r3=2000
r3=125
r=3{125}=5
Since h=1000πr2, h=1000π(5)2=40π
Therefore, the ratio of h to r is 40π5=8π
(Note: Since π is less than 4, the new ratio is greater than 2 to 1)

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Natalie Yamamoto
Answered 2021-12-20 Author has 22 answers
Step 1
Total area of the can=area of (top+bottom)+lateral area
lateral area 2πrh without waste
area of base (considering that you use 2r square) is 4r2
area of bottom ( for same reason ) 4r2
Then Total area =8r2+2πrh
Now can volume is 1000=πr2h
h=1000πr2
And A(r)=8r2+2πr(1000)πr2
A(r)=8r2+2000r
Taking derivatives both sides
A(r)=16r2000r2
If A(r)=0
16r2000r2=0
16r32000r2=0
16r32000=0
r3=125
r=5cm and h=1000πr2
h=10003.14×25
h=12.74cm
ratio hr=12.745
hr=2.55

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