The width of a rectangle is 20m less than twice its length. If the perimeter of the rectangle is more than 56m, find the minimum dimensions of the rectangle if each dmension in meters is an integer.
length ----x m
width ---- 2x-20 m
2x + 2(2x-20) > 56
6x - 40 > 56
6x > 96
x > 16
min length is 17m, min width is 14 m
note, a length of 16 and a width of 12 would give you an exact perimeter of 56, but it said it had to be greater than 56 , and the sides had to be integers.
Let's assume the length of the rectangle is L meters.
According to the given information, the width is 20 meters less than twice its length.
So, the width of the rectangle is 2L - 20 meters.
The perimeter of a rectangle is given by the formula:
Perimeter = 2 * (Length + Width)
Given that the perimeter is more than 56 meters, we can write the inequality:
2 * (L + (2L - 20)) > 56
Simplifying the inequality:
2 * (L + 2L - 20) > 56
2 * (3L - 20) > 56
6L - 40 > 56
6L > 96
L > 16
Since the dimensions of the rectangle need to be integers, the minimum value for the length L is 17.
So, the minimum dimensions of the rectangle are:
Length = 17 meters
Width = 2L - 20 = 2 * 17 - 20 = 14 meters
To find the minimum dimensions of the rectangle, we need to set up an equation based on the given information.
Let's assume the length of the rectangle is "x" meters. According to the problem, the width is 20 meters less than twice the length, so the width can be expressed as (2x - 20) meters.
The formula for the perimeter of a rectangle is given by P = 2(length + width). We are given that the perimeter is more than 56 meters, so we can write the equation as:
2(x + (2x - 20)) > 56
Simplifying the equation, we get:
2(3x - 20) > 56
6x - 40 > 56
6x > 96
x > 16
Since we need the dimensions to be integers, the minimum value for x will be 17. Let's substitute this value back into the equation to find the width:
Width = 2x - 20 = 2(17) - 20 = 34 - 20 = 14
Therefore, the minimum dimensions of the rectangle are 17 meters for the length and 14 meters for the width.