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3 years ago

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A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 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 2706 incubed, what were the original dimensions of the piece of metal?

1 answer:

Alexandra [31]3 years ago

7 0

Answer:

length=53 in

width= 23 in

Step-by-step explanation:

First, we know that each corner has now a square of 6 in long, so the heigth of the box will be of this size (as you can see in the picture attached). Then, the length and the width of the box would be the same as the piece, but with 12 in less. Therefore the equations to solve the problem are:

$h=6 in\\w=x-12\\l=x+30-12\\l=x+18\\\\V: volume of the box\\V=6*(x-12)*(x+18)\\2706=6x^{2}+36x-1296\\0=6x^{2}+36x-4002\\\\$

We factorize the 6 and we get:

$0=x^{2}+6x-667\\$

Finally we solve for x, and we get the initial dimensions of the piece of metal:

$x=23$ $l=30+x=30+23=53 in\\w=x= 23 in$

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