algebra word problem

Question

a rectangular piece of metal is 25 in longer than it is wide. squares with sides of 5 in are cut from each corner and the flaps are folded upward to form an open box. if the volume of the box is 1120 what were the original dimensions of the metal?

Answer 1

Jade,

 

Let the original width of the metal = w, then the original length of the metal will be w+25.

 

Now subtract from the length and width the 5" square corners so the folded width will be:  (w-10)

The folded length will be:  (w+25-10) = (w+15)  (we subtracted 5" from each corner/end)

 

The height of the box will be 5" so with the volume we can write:  (w-10)(w+15)(5) = 1120  

Divide both sides by five:

 

(w-10)(w+15) = 224 ⇒ w²-10w+15w-150 = 224 ⇒ w²+5w-150 = 224

 

setting our quadratic formulas = 0:

 

w²+5w-374 = 0

 

This can be solved most easily using the quadratic formula, and ignoring the negative root:

 

w = 17"

l = 17"+25" = 42"

 

So our original sheet of metal measured 17" x 42"

 

Check: (17-10)(42-10)(5) = 7x32x5 = 1120