Dimensions of a Box

Question

A rectangular shipping crate is in the form of a cube. The width is decreased by 2 m while the
length is increased by 2m and the height is increased by 3 m. The new crate has a volume of
168 m^3. Find the dimensions of the new crate.

Answer 1

Let x be the side dimension of original crate

 

New Crate Dimensions:  Length: x = 2,  Width: x - 2,  Height: x + 3

 

(x - 2)(x + 3)(x + 2) = 168

 

(x²  + x - 6)(x +2) = 168

 

x³ + 3x² - 4x -180 = 0

 

Using a Graphing Calculator: You would see that x = 5  is the only Real solution

 

Therefor the Original Cube had an Edge length of 5

 

Dimensions for the New Crate are: Length is 7, Height is 8, and Width is 3;  (7)(8)(3)= 168

Answer 2

Since the crate began as a cube, we know all the sides have the same length, we will call that length "L".

 

The volume of the crate will be               length x width x height. 

 

For the original crate, that would be           (L)     x    (L)   x    (L)

 

The width was decreases by 2 so we have:  (L)    x    (L-2)  x   (L)

 

The length is increased by 2 so we have: (L+2)  x   (L-2)   x   (L)

 

The height is increased by 3 so we have:  (L+2)  x   (L-2)   x  (L+3)

 

 

We know the new volume is 168, so

 

 (L+2) x (L-2) x (L+3) = 168

 

  To solve this we nee to  multiply everything out we will start with the Length and width. (I will use the FOIL method, multiplying the First term inside each parentheses, then the Outside terms, then the Inside terms then the Last two term in each parentheses. If you have any questions about this method please let me know)

 

(L+2) (L-2) = (L² - 2L + 2L - 4) = (L² - 4)

 

So we have :(L² - 4)x (L+3)   =168

 

Now we need to multiply the rest using the same FOIL method.

 

(L²-4) x (L+3)= (L³ + 3L² - 4L - 12) = 168

 

Now we solve for L

L³ + 3L² - 4L - 12 = 168

 

Subtract 168 from both sides

L³ + 3L² - 4L - 180 = 0

 

Factor:

 (L-5)(L²+8l+36)=0

 

**Factoring trinomial can be tricky, let me know if you would like an explanation of how I factored this)**

 

We have two thing multiplied together to get 0 so one of them must be 0.

                        (L-5)=0       or        (L²+8L+36)=0 ---> use quadratic equation to solve

  

 

 which means       L=5          or        L=√(-20)-4 ---> NOT a real number so we can eliminate this answer.

 

So  L= 5  and you can use the first equations (L+2), (L-2), (L+3) to determine Length, width and height respectively.

 

Length=7

Width=3

Height=8

 

Please let me know if you have any questions.

 

Answer 3

Hi Tish.

 

Since the original crate is a cube, that tells us the length, width and height are equal.

 

LxWxH = V

 

Now for the new crate,

 

(L+2)(W-2)(H+3) = 168

 

Since L=H=W we can substitute in the equation above and just solve for one of the variables.  I'm going to solve for L

 

(L+2)(L-2)(L+3) = 168

(L²-4)(L+3) = 168

L³+3L²-4L-12 = 168 which becomes

L³+3L²-4L-180 = 0

 

Working with this cubic we can factor it,

(L-5)(L²+8L+36) = 0

L-5 = 0 so L=5 is the only real solution.  If we try to factor the quadratic, we do not get real solutions.

 

So the original length of the box was 5m

 

Now the length is increased by 2

The width is decreased by 2

and the height is increased by 3

Leaving us with the following dimensions.

(5+2) x (5-2) x (5+3)

7 x 3 x 8

 

Which gives us a volume of 168 m³

 

The dimensions of the new crate are

Length 7m

Width 3m

Height 8m