Answer
a) and b) see image
c) V = length * width * height
d) length = 3 - 2x
width = 3 - 2x
height = x
Volume = ( 3 - 2x )( 3 - 2x )( x )
e) V ( x ) = x$( 3 - 2x )^{2}$
f) Maximum volume: 2 ft³
Work Step by Step
a) and b)
Given:
Each side of the initial square is 3 ft, so the diagram should start as a square.
Each corner has a square with an unknown side length ( x ) removed to fold the sides up, so every corner should have equal squares with sides labelled x removed.
The removed corner squares are the height the box goes vertically, so the side length of the removed squares is height.
Unknown: The maximum volume of the box, so when f ' ( c ) = 0 and c > all other x-values on the interval.
c) Volume of a rectangular prism = length * width * height
d) length = 3 - 2x
width = 3 - 2x
height = x
V = length * width * height
V = ( 3 - 2x ) * ( 3 - 2x ) * ( x )
e) V ( x ) = x$( 3 - 2x )^{2}$
f) To find absolute maximums or minimums find V ' = 0
Make less work in finding $\frac{dV}{dx}$ ( the derivative ) by simplifying first.
V ( x ) = x$( 3 - 2x )^{2}$
V ( x ) = x( 9 - 12x + 4x² )
V ( x ) = 9x - 12x² + 4x³
$\frac{dV}{dx}$ = $\frac{d}{dx}$ [ 9x - 12x² + 4x³ ]
$\frac{dV}{dx}$ = 9 - 24x + 12x²
0 = 9 - 24x + 12x²
0 = 3( 3 - 8x + 4x² )
0 = 4x² - 8x + 3
0 = ( 2x - 3 )( 2x - 1 )
0 = ( 2x - 3 )
$\frac{3}{2}$ = x
V ( $\frac{3}{2}$ ) = ( $\frac{3}{2}$ )$( 3 - 2( \frac{3}{2} ) )^{2}$
V ( $\frac{3}{2}$ ) = ( $\frac{3}{2}$ )$( 3 - 3 )^{2}$
V ( $\frac{3}{2}$ ) = ( $\frac{3}{2}$ )$( 0 )^{2}$ = 0
The minimum area of the box is 0 ft³, the other x-value could be the maximum.
0 = ( 2x - 1 )
$\frac{1}{2}$ = x
V ( $\frac{1}{2}$ ) = ( $\frac{1}{2}$ )$( 3 - 2( \frac{1}{2} ) )^{2}$
V ( $\frac{1}{2}$ ) = ( $\frac{1}{2}$ )$( 3 - 1 )^{2}$
V ( $\frac{1}{2}$ ) = ( $\frac{1}{2}$ )$( 2 )^{2}$
V ( $\frac{1}{2}$ ) = ( $\frac{1}{2}$ )( 4 )
V ( $\frac{1}{2}$ ) = 2
The maximum volume is 2 ft³.
