Word Problem

Question

A rectangular piece of metal is 30 in longer than it is wide. Squares with 6 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 3354³, what were the original dimensions of the piece of metal?

Answer 1

In the beginning:

 

w = width

w + 30 = length

 

After cutting:

w - 12 = width

w + 30 - 12 = length, or w + 18

 

(w - 12)(w + 18)(6) = 3354

(w - 12 )(w + 18) = 559

w² + 6w - 216 = 559

w² + 6w - 775 = 0

(w + 31)(w - 25) = 0

(-31, 25}

 

The width is 25 and the length is 55.

 

Check:

(25 - 12)(55 - 12)(6)

(13)(43)(6)

3354

 

 

Answer 2

Let L be the length of the rectangle and W be the width of the rectangle.  Then L=30+W.

 

Now, let x be the length of the horizontal side of the rectangle when the squares are cut out from the corners.  Then, x+2*6=L, because there is a length of 6 cut out twice, one from the top and one from the bottom.  Similarly, let y be the length of the side of the vertical side of the rectangle after the squares are cut out from the corners.  Then, y+2*6=W.

 

The volume of the box is length*width*height.  The length is x, the width is y, and the height is 6.  We know that x*y*6=3354. Solve for x.  x=3354/(6y)=559/y.

 

So, plug this into x+2*6=L, to get 559/y+12=L.  We know that L=30+W, so 559/y+12=30+W.  But we know y+2*6=W.  So 559/y+12=30+y+12.

 

Solve this for y.  Multiply both sides by y to get 559=30y+y^2.  This gives y=13.  Then plug this into x=559/y.  Then x=43.

 

Use x+12=L to solve for L.  Use y+12=W to solve for W.