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?

h / r = 2.55

Step-by-step explanation:

Area of a can:

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 = 1000/πr²

And A (r) = 8r² + 2πr (1000) / πr²

A (r) = 8r² + 2000/r

Taking derivatives both sides

A' (r) = 16 r - 2000/r²

If A' (r) = 0 16 r - 2000/r² = 0

(16r³ - 2000) / r² = 0 16r³ - 2000 = 0

r³ = 125

r = 5 cm and h = (1000) / πr² h = 1000 / 3.14 * 25

h = 12,74 cm

ratio h / r = 12.74/5 h / r = 2.55