You are designing a 1000 cm^3 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 beA = 8r^2 + 2pi rhWhat is the ratio now of h to r for the most economical can?

Answer:  h /r  =  2.55Step-by-step explanation:Area of a can:  Total area of the can = area of (top + bottom)  + lateral arealateral area  2πrh    without wastearea of base (considering that you use 2r square)  is 4r²area of bottom      ( for same reason )                           4r²Then Total area  =  8r² + 2πrhNow can volume is   1000 = πr²h        h = 1000/πr²And  A(r)  =   8r² + 2πr(1000)/πr²A(r)  =   8r² + 2000/rTaking derivatives both sidesA´(r)   =  16 r    - 2000/r²If A´(r)  =  0              16 r    - 2000/r²   = 0(16r³  - 2000)/ r²  =  0        16r³  - 2000  =  0r³  =  125       r = 5 cm      and      h  = ( 1000)/ πr²      h  = 1000/ 3.14* 25h = 12,74  cmratio   h /r  =  12.74/5         h /r  =  2.55