the material for a can is cut from sheets of metal. the cylindrical sides are fo
ID: 2838580 • Letter: T
Question
the material for a can is cut from sheets of metal. the cylindrical sides are formed by bending rectangles: these rectangles are cut fromt he sheet with little or no waste. But if the top and bottom discs are cut from squares of side 2r, this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. if this is the case, show that the amount of metal used is minimized when h/r=8/pi or appx 2.55
Can you please give a detailed answer, so that I'll be able to understand it? Thank you!
Explanation / Answer
Material = 2?rh (the side) + 2(2r)2 = 2?rh + 8r 2 V = ?r2h, or h = V/(?r2) Thus, we will minimize 2?rV/(?r2 ) + 8r2 = 2V/r + 8r2 f'(r) = -2V/r2 + 16r Then, 2V/r2 = 16r or, crossmultiplying, 16r 3 = 2V, or r 3 = 1/8 V or r = 1/2 V 1/3 Note that the second derivative 4V/r3 + 16, is positive for all positive r, so this is a minimum. Then, h = V/(?r2 ) = V/(? (1/2V1/3)2 ) = V/(?*1/4V2/3) = 4 V1/3/? and h/r = 4V1/3/?/(1/2V 1/3 = 8/? ? 2.55
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