Construction word problem

Question

A rectangular container measuring 1 foot by 2 feet by 4 feet is covered with a layer of lead shielding of uniform thickness.

1.) find the volume of lead shielding V as a function of the thickness x (in feet) of the shielding

2.) find the thickness of the lead shielding to three decimal places if the colkme of the shielding is 3 cubic feet

 

I'm having trouble getting for both 1.) V=28x+28x^2+8x^3

2.) 0.0097ft

Answer 1

x is the doubled thickness of the shielding in feet

(Total volume including shielding) minus (Volume of container)

(1+x)(2+x)(4+x) - (1)(2)(4)

x³ + 7x² + 14x + 8 - 8 = Vs
x³ + 7x² + 14x = Vs

Solve for thickness if volume is 3 ft³
x³ + 7x² + 14x = 3

x³ + 7x² + 14x - 3 = 0

It is clear this is not factorable by any positive rational root, so using Synthetic Division and the Location Principle I found 0.1948 ft to produce the closest, three digit, value for x/2. (My trial values were 0.2, 0.19, 0.193, 0.194, 0.195, 0.1948, 0.1947)  

 

Since 0.1948 is the doubled thickness, 0.0974 ft Is the thickness of the shielding.

 

 

 

Answer 2

On the original rectangular container there are 6 sides

The will be... 2 sides at 1 ft by 2 ft, 2 sides at 1 ft by 4 ft, 2 sides at 2 ft by 4 ft

 

1)  The volume of the shielding will be 6 found as length*width*height of 6 solid

      rectangular objects...

 

The volume of lead shielding is:

 

V = 2(1*2*x) ft³ + 2(1*4*x)ft³ + 2(2*4*x)ft³

= 4x ft³ + 8x ft³ + 16x ft³

= 28x ft³

 

2) If the volume of the shielding is 3 cubic ft then

 

    28x = 3

     x = 3/28

     x = .107 ft