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 cubed3โ,what were the original dimensions of the piece ofโ metal?

What is the original width? in

yojaymarojas
May 13, 2017

#1**+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 = w^{2} + 5w - 150

0 = w^{2} + 5w - 300

0 = (w + 20)(w - 15)

w = -20 or w = 15

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

hectictar
May 13, 2017

#1**+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 = w^{2} + 5w - 150

0 = w^{2} + 5w - 300

0 = (w + 20)(w - 15)

w = -20 or w = 15

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

hectictar
May 13, 2017