Solve the problem.
-A rectangular piece of tin has an area of 496 square inches. A square of 2 inches is cut from each corner, and an open box is made by turning up the ends and sides. If the volume of the box is 648 cubic inches, what were the
Original dimensions of the piece of tin?
A)18 in. by 33 in.
B)12 in. by 25 in.
C)16 in. by 31 in.
D)14 in. by 29 in.

Question

Quizplus · Accepted Answer

Let the original dimensions of the tin be x and y. 
After cutting the squares from all four corners, the dimensions of the base of the open box become (x-4) by (y-4). 
The height of the box is 2 inches (the height of the removed square). 
Therefore, the volume of the box is (x-4)(y-4)(2). 
We are given that this volume is 648 cubic inches, so we have the equation: 
(x-4)(y-4)(2) = 648 
Expanding this equation and simplifying, we get: 
xy - 8x - 8y + 32 = 324 
xy - 8x - 8y - 292 = 0 
Using the given area of the tin (xy = 496), we can substitute for xy in this equation: 
496 - 8x - 8y - 292 = 0 
204 = 8x + 8y 
51/2 = x + y 
Now we need to find the dimensions of the original tin, given that x+y = 51/2 and xy = 496. 
We can solve for y in terms of x using the first equation: 
y = 51/2 - x 
Substituting into the second equation and simplifying, we get a quadratic equation: 
x^2 - (51/2)x + 496 = 0 
This factors as: 
(x-16)(x-31) = 0 
Therefore, x = 16 or x = 31. 
If x = 16, then y = 51/2 - 16 = 19/2. But this does not give us an integer answer for the area, so we must discard it. 
If x = 31, then y = 51/2 - 31 = 11/2. This does give us an integer answer for the area (31*11 = 341), so the original dimensions of the tin were 31 inches by 11 inches. The answer is therefore choice C. Let the original dimensions of the tin be $x$ and $y$. When 2 inch squares are cut from each corner, the dimensions of the base of the box will be $x-4$ and $y-4$. The height of the box will be 2 inches.

The volume of the box can be expressed as:

$$(x-4)(y-4)(2) = 648$$

Simplifying this equation, we get:

$$(x-4)(y-4) = 324$$

We also know that the area of the tin is 496 square inches:

$$xy = 496$$

We can solve these two equations to find that $x=31$ and $y=16$. Therefore, the original dimensions of the tin were 31 inches by 16 inches.

The answer is choice C.

Solve the Problem. -A Rectangular Piece of Tin Has an Area of 496