Question 1197396
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A rectangular piece of metal is 15 in longer than it is wide. Squares with sides 3 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 1092 in​^3, what were the original 
dimensions of the piece of​ metal? What is the original​ width?
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Let x be the original width of the piece of metal.

Then the original length is (x+15) inches.


After cutting the squares and folding the dimensions of the base of the open box are (x-6) and (x+9) inches 
(by 2*3 = 6 inches less than the original dimensions).


Therefore, the volume equation is

    3*(x-6)*(x+9) = 1092,

or, equivalently

    (x-6)*(x+9) = 1092/3 = 364.


Simplify and find x

    x^2 - 6x + 9x - 54 = 364

    x^2 + 3x - 418 = 0

    {{{x[1,2]}}} = {{{((-3) +- sqrt((-3)^2 - 4*1*(-418)))/2}}} = {{{((-3) +- sqrt(1681))/2}}} = {{{(-3 +- 41)/2}}}.


It gives two roots, one negative and one positive.


We reject the negative root and accept the positive one  x = {{{((-3) + 41)/2}}} = 19.


<U>ANSWER</U>. The original width was 19 inches.

        The original length was 19 + 15 = 34 inches.


<U>CHECK</U>.  The volume is  3*(19-6)*(34-6) = 3*13*28 = 1092 in^3, which is precisely correct.
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Solved.