+0

# Algebra question

0
147
1
+52

A rectangular piece of metal is 20 in longer than it is wide. Squares with sides 4 in long are cut from the four coners and the flaps are folded upward to form an open box. If the volume of the box is 1904 in^3, what were the original dimensions of the piece of metal?

What is the width?   in

What is the original length?    in

yojaymarojas  May 13, 2017
Sort:

#1
+79794
+1

Let the original width be  = W

And the original length = W + 20

When the squares are cut  the   dimensions of the box  = (W - 8) in, (W + 20 - 8) in and 4 in

And we have that

(W - 8) ( W + 20 - 8) (4)  =  1903

(W - 8) ( W + 12) (4)  =  1903     simplify

4W^2  + 16W - 384 = 1903

4W^2 + 16W - 2287  = 0

Using the quadratic formula to solve this ( and taking the positive answer) the  solution is that W ≈ 21.995 in

And the length  ≈  21.995 + 20 ≈  41.995 in

CPhill  May 14, 2017

### 23 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details