Solve the problem.
-A rectangular piece of tin has an area of 612 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 832 cubic inches, what were the
Original dimensions of the piece of tin?
A) 13 in. by 30 in.
B) 19 in. by 38 in.
C) 17 in. by 36 in.
D) 15 in. by 34 in.

Question

Quizplus · Accepted Answer

Let the original length and width be x and y respectively.
After cutting 2 inches from each corner, new length and width will be (x-4) and (y-4) respectively.
Given, volume of the box = 832 cubic inches
Volume of the box = (length * width * height)
⇒ 832 = (x-4)(y-4)(2)
⇒ 416 = (x-4)(y-4)  ......(1)
Also, area of the rectangular tin = 612 square inches
⇒ xy = 612   ......(2)
We need to find the original dimensions of the piece of tin i.e. x and y.
From equation (1), we know that (x-4)(y-4) = 416
Let's find the factors of 416:
416 = 1 * 416
416 = 2 * 208
416 = 4 * 104
416 = 8 * 52
416 = 13 * 32
416 = 16 * 26
416 = 26 * 16
416 = 32 * 13
416 = 52 * 8
416 = 104 * 4
416 = 208 * 2
416 = 416 * 1
Out of these factor pairs, (32,13) and (26,16) are the only pairs whose difference is 4.
So, (x-4) = 32 and (y-4) = 13  OR  (x-4) = 26 and (y-4) = 16
From equation (2), xy = 612
Substituting the values of (x-4) and (y-4) in the above equation, we get:
xy = 612
(32+4)(13+4) = 612  OR  (26+4)(16+4) = 612
36*17 = 612  OR  30*20 = 612
612 = 612  OR  600 ≠ 612
Therefore, the only feasible solution is (x-4) = 32 and (y-4) = 13
Hence, the original dimensions of the piece of tin are x = 36 and y = 17
Therefore, the answer is C.

Solve the Problem. -A Rectangular Piece of Tin Has an Area of 612