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A rectangular piece of metal is 25 in longer than it is wide. Squares with sides

5in 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 750in cubed3​,what were the original dimensions of the piece of​ metal?

 

What is the original width?   in

 May 13, 2017

Best Answer 

 #1
avatar+9479 
+4

Imagine folding this up like it says in the problem...we can see that

 

height of box = 5

width of box = w - 5 - 5 = w - 10

length of box = (w + 25) - 5 - 5 = w + 15

 

volume of box = (height)(width)(length)

750 = (5)(w - 10)(w + 15)

750/5 = (w - 10)(w + 15)

150 = w2 + 5w - 150

0 = w2 + 5w - 300

0 = (w + 20)(w - 15)

w = -20     or     w = 15

 

So...the original width must be 15 in.   smiley

 May 13, 2017
 #1
avatar+9479 
+4
Best Answer

Imagine folding this up like it says in the problem...we can see that

 

height of box = 5

width of box = w - 5 - 5 = w - 10

length of box = (w + 25) - 5 - 5 = w + 15

 

volume of box = (height)(width)(length)

750 = (5)(w - 10)(w + 15)

750/5 = (w - 10)(w + 15)

150 = w2 + 5w - 150

0 = w2 + 5w - 300

0 = (w + 20)(w - 15)

w = -20     or     w = 15

 

So...the original width must be 15 in.   smiley

hectictar May 13, 2017

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