Modeling Word Problem -- Quadratic Functions #57

Question

a piece of cardboard is twice as long as it is wide. it is to be made into a box with an open top by cutting 2-in. squares from each corner and folding up the sides. let x represent the width (in inches) of the original piece of cardboard. 

 

a). represent the length of the original piece of cardboard in terms of x

b). what will be the dimensions of the bottom rectangular base of the box? give the restrictions on x

c). determine a function V that represents the volume of the box in terms of x

d). for what dimensions of the bottom of the box will the volume be 320 in.³

e). find the values of x (to the nearest tenth of an inch) f such a box is to have colume between 400 and 500 in.³

Answer 1

1.  Let x=width of the cardboard, then 2x=length.

 

2.  When we cut out the 2" squares and fold the sides up, the dimensions of the bottom of the box is x-4 and 2x-4.  the restrictions on x are INF < x < 4.

 

3.  V=(x-4)(2x-4)(2) = 4x² - 24x +32

 

4.  320 = 4x² - 24x + 32

      80 = x² -6x + 8

      x² - 6x -72 = 0

     (x-12)(x+6) = 0

   x = 12 or -6, -6 is excluded from the domain.

 

Therefore the dimensions of the bottom of the box will be 8 and 20

 

5.  400 = 4x² -24x + 32

     100 = x² -6x + 8

       x² - 6x - 92 = 0

 

Solve with the quadratic formula to get your lower boundary

 

500 = 4x² - 24x + 32

125 = x² - 6x + 8

x² - 6x - 117 = o

 

Solve with the quadratic formula to get the upper boundary