MATH SOLVE

4 months ago

Q:
# 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?

Accepted Solution

A:

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