Solve the problem.
-A rectangular piece of tin has an area of 592 square inches. A square of 4 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 928 cubic inches, what were the
Original dimensions of the piece of tin?
A) 8 in. by 25 in.
B) 20 in. by 41 in.
C) 16 in. by 37 in.
D) 12 in. by 33 in.

Question

Quizplus · Accepted Answer

Let the original dimensions of the tin be $x$ and $y$ inches. After cutting 4-inch squares from each corner and folding, the new dimensions become $x-8$ and $y-8$ inches, with a height of 4 inches (the side of the square cut out). The volume of the box is given by $V = (x-8)(y-8)(4)$. We know $V = 928$ cubic inches and the area of the original piece of tin is $xy = 592$ square inches.Solving the system:1. $xy = 592$2. $(x-8)(y-8)(4) = 928$From equation 2, simplifying gives $(x-8)(y-8) = 232$. Substituting $x = \frac{592}{y}$ from equation 1 into equation 2 and solving for $y$ (or vice versa), we find that the dimensions that satisfy both the area and volume conditions are those of choice C, which are 16 inches by 37 inches.

Solve the Problem. -A Rectangular Piece of Tin Has an Area of 592