First things first:
width --- s
length ---- 2s
2s^2 = 800
s^2 = 400
s = 20
so the piece of metal is 20 by 40
let the side of the square to be cut out be x
so the width is 20-2x
the length is 40-2x
the height is x
Volume = x(20-2x)(40-2x)
= 2x^3 - 120x^2 + 800x
d(Volume)/dx = 6x^2 - 240x + 800 = 0 for a max/min of Volume
3x^2 - 120x + 400 = 0
use the formula to solve for x
remember that x must be between 0 and 10 or else you get a negative side.
(hope you get appr 3.67
You are given a piece of sheet metal that is twice as long as it is wide an has an area of 800m^2. Find the dimensions of the rectangular box that would contain a max volume if it were constructed from this piece of metal by cutting out squares of equal area at all four corners and folding up all the sides. The box will not have a lid.
5 answers
messed up in my expansion
volume should have been
4x^3- 120x^2 + 800x
and V' = 12x^2 - 240x + 800 = 0
3x^2 - 60x + 200=0
to get x = 4.23
volume should have been
4x^3- 120x^2 + 800x
and V' = 12x^2 - 240x + 800 = 0
3x^2 - 60x + 200=0
to get x = 4.23
Why is the max value of x 10?
otherwise, the width which is 20 meters would be less than 0 after you remove 10 meters twice (one x per corner).
Therefore, we cannot have a rectangular plot of land with a width of 20 meters and two 10 meter cuts without one of the sides being a length less than 0.
1 answer