SOLUTION: 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

Question 1197396: 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?
Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
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?
~~~~~~~~~~~~~~~~~~~~~~~~~

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 = = = . It gives two roots, one negative and one positive. We reject the negative root and accept the positive one x = = 19. ANSWER. The original width was 19 inches. The original length was 19 + 15 = 34 inches. CHECK. The volume is 3*(19-6)*(34-6) = 3*13*28 = 1092 in^3, which is precisely correct.

Solved.

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
x, the width of the rectangle piece
x+15, length of the rectangle piece
3, HEIGHT of the box formed

The base area becomes .

VOLUME is .
This may be solvable through checking factoring.

--------try to look for two factors for 364 which differ by 15. Worth a try!

-
---------solve either for x.
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

x = width of sheet
x+15 = length

After cutting the 3-inch squares out of the corners and folding the flaps up, the dimensions of the open box are
width: x-2(3) = x-6
length: (x+15)-2(3) = x+9
depth: 3

The volume is 1092 cubic inches:

It looks as if that quadratic is going to be hard to factor, so we could use the quadratic formula. However, if the problem is well designed, the answer should be an integer, so let's try to do the factoring.

We need two numbers whose product is 418 and whose difference is 3.

418 = 2(209) = 2(11)(19) = (22)(19)

So

or

Clearly the negative answer makes no sense in the problem, so

ANSWER: the original width is x = 19 inches

CHECK: volume = 3(x-6)(x+9) = 3(13)(28) = 1092

Note that if a formal algebraic solution is not required, you can solve the problem with a bit of logical reasoning and some simple arithmetic.

The volume is 1092, and the depth is 3; so length times width is 1092/3 = 364.

So now we need two integers whose difference is 15 and whose product is 364:

364 = 4(91) = 4(7)(13) = 28*13

The width of the box is 13 inches; so before the 3-inch squares were cut out of the corners, the original width was 13+6 = 19 inches.

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