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

really really sorry if i'm

-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x

1. supplementary angles

2. acd and bcd

step-by-step explanation: