Question 154297This question is from textbook
: Production Design
A candy company makes a washer –shaped candy (a candy with a hole in it);because of increasing cost ,the company will cut the volume of candy in each piece by 22%.To do this, the firm will keep the same thickness and outer radius, but will make the inner radius larger. At present the thickness is 2.1mm,the inner radius is 2mm,and the outer radius is 7.1mm.Find the inner radius of the new-style candy(hint :the volume V of a solid disk is Pi r^2h , where r is the radius and h is the thickness of the disk)
This question is from textbook
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A candy company makes a washer –shaped candy (a candy with a hole in it);
because of increasing cost ,the company will cut the volume of candy in each piece by 22%.
To do this, the firm will keep the same thickness and outer radius, but will make the inner radius larger.
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At present the thickness is 2.1mm,the inner radius is 2mm,and the outer radius is 7.1mm.
Original volume as if there were no hole: (pi)7.1^2*2.1 = 105.861pimm^3
Volume of the hole: pi*2*2.1 = 4.2pimm^3
Therefore volume of the candy portion: (105.861-4.2)pimm^3 = 101.661pimm^3
Find the inner radius of the new-style candy(hint :the volume V of a solid disk is Pi r^2h , where r is the radius and h is the thickness of the disk)
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If you reduce the volume by 22%
EQUATION:
0.22*101.661pimm^3 = 22. 365420mm^3 of candy
Replace that volume with "hole" to get a total hole of (4.2+22.365420)pi mm^3
= 26.565420 pi mm^3
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Find the radius of this new hole:
V = (pi)r^2h
22.565420 pi mm^3 = pi*r^2 *2.1
r^2 = 22.565420/2.1 = 10.650200
r = 3.26344 mm
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Cheers,
Stan H.
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