Find solutions to inequalities in one variable and for a specified variable - Essay Example

Share a popular formula or application that has more two or more variables. Provide a specific solution to the formula or application that you posted and share with your class. Why did you choose this specific formula or application? In finding the perimeter of a rectangle, the popular formula used is P = 2L + 2W where ‘P’ stands for the perimeter whereas variables ‘L’ and ‘W’ refer to the length and the width of the rectangle, respectively. Application #1: The length of a rectangular lot is 147 meters while its width is 56 meters.

Find the lot’s perimeter. Solution: Upon substitution of the given values into the formula P = 2L + 2W— P = 2*(147 m.) + 2*(56 m.) = 294 m. + 112 m. P = 406 metersApplication #2: A newly bought rectangular lawn needs to be fenced with 450 yards of fencing material. If the length must be 3 yards more than twice the width, what should be the length and the width of the lawn?Solution: Let x = width (in yards) so that 2x + 3 = length (in yards) according to the problem statement.

Then substituting these expressions into P = 2L + 2W— 450 = 2*(2x + 3) + 2x 450 = 4x + 6 + 2x Combining similar terms: 450 - 6 = 4x + 2x --- 444 = 6x Dividing each side by 6: x = 74 yardsSo this makes 2x + 3 = 2*(74) + 3 = 151 yards Therefore, width = 74 yards and length = 151 yardsI chose P = 2L + 2W because I have observed several practical applications can be carried out through it when it comes to finding dimensions of a rectangular figure.

It is of huge significance, I believe, to construction industry where taking measurements in relation to perimeter is often necessary.