Question

An open box is to be made from a rectangular piece of tin by cutting 3-inch squares out of the corners and folding up the sides. The length of the piece of tin is twice the width. The volume of the box will be 60 cubic inches. Find the dimensions of the original rectangular piece of tin

Answer 1

Answer:

Width of the Rectangular Tin=8 inch

Length of the Rectangular Tin= 16 inch

Step-by-step explanation:

Let the Width of the Rectangle=W

The length of the piece of tin is twice the width, Length = 2W

Since Squares of 3 inch are cut from all four corners of the rectangle

Length of the box = 2W-(3+3)=(2W-6) inches

Breadth of the Box = W-(3+3)=(W-6) inches

Height = 3 inches

Volume of the box = 60 cubic inches

Now, Volume of a cuboid=lbh

3(2W-6)(W-6)=60

Divide both sides by 3

(2W-6)(W-6)=20

Expanding the brackets

$(2W-6)(W-6)=20\\2W^2-12W-6W+36=20\\2W^2-18W+36-20=0\\2W^2-18W+16=0$

Factorizing

$2W^2-16W-2W+16=0\\2W(W-8)-2(W-8)=0\\(W-8)(2W-2)=0\\W-8=0 , 2W-2=0\\W=8, 1$

Since the Width cannot be less than 6,

Width of the Rectangular Tin=8 inch

Length= 2 X 8 = 16 inch