Question #2a36d

Step 1
Imagine a cube box.

Based on this image, we can see that there are #6# sides.

To find the maximum possible volume for the box, we must determine two things - the area of one side and the side length. Knowing that there is #128color(white)(i)cm^2# of material, we can divide #128color(white)(i)cm^2# by #6# to give us the area for one side of the box.

#128color(white)(i)cm^2-:6#
#=64/3color(white)(i)cm^2#

Step 2
We can now use the calculated value to determine the side length of the box. Recall that the area of a square is given by:

#color(blue)(|bar(ul(color(white)(a/a)A=s^2color(white)(a/a)|)))#

#ul("where:")#
#A=#area
#s=#side length

Now that we have determined a relationship between area and side length, we can plug our values into the formula to determine the side length.

#A=s^2#

#s=+-sqrt(A)#

#s=+-sqrt(64/3)#

Note: Since a measurement cannot be negative, the only valid answer is #8/sqrt(3)# and not #-8/sqrt(3)# !

#s=8/sqrt(3)#

Step 3
Since we have now determined the side length, we can use the formula for volume of a cube to determine its maximum possible volume:

#color(blue)(|bar(ul(color(white)(a/a)V=s^3color(white)(a/a)|)))#

#ul("where:")#
#V=#volume
#s=#side length

Plugging in the values,

#V=s^3#

#V=(8/sqrt(3))^3#

#V=512/(3sqrt(3))#

#V~~color(green)(|bar(ul(color(white)(a/a)color(black)(98.53color(white)(i)cm^3)color(white)(a/a)|)))#