SOLUTION: A rectangular pen is 3 feet longer than its wide. Give the possible values for the width W of the pen if its area must be between 154 and 460 inclusively. I'm bad enough at geomet

Algebra ->  Rectangles -> SOLUTION: A rectangular pen is 3 feet longer than its wide. Give the possible values for the width W of the pen if its area must be between 154 and 460 inclusively. I'm bad enough at geomet      Log On


   

Question 453088: A rectangular pen is 3 feet longer than its wide. Give the possible values for the width W of the pen if its area must be between 154 and 460 inclusively.
I'm bad enough at geometry, but I actually thought I was getting the hang of finding areas until I saw this problem. Can someone help please? I've never even seen anything like this before. Thanks
Answer by Sarpi(32) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= the width and therefore length= 3+x
Area= L*W = (3+x)*x
So by simple simplification, the area has the equation,x%5E2%2B3x
We are bounded to find values of width within the areas of 154 and 460. Which means this equation must hold true, 154%3Cx%5E2%2B3x%3C460
we solve two equations:
x%5E2%2B3x%3C460...... eqn 1
x%5E2%2B3x%3E154 ..... eqn 2
The two equations have a quadratic system that is ax%5E2%2Bbx%2Bc=0
Eqn 1: x%5E2%2B3x-460%3C0 using this formula x%3C%28-b%2B-sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29
will give x%3C-23+and+x%3C20 that is: x%3C%28-3%2B-sqrt%283%5E2-4%2A1%2A-460%29%29%2F%282%2A1%29
Eqn 2: x%5E2%2B3x-154%3E0 also with the formula gives x%3E-14+and+x%3E11 that is: x%3E%28-3%2B-sqrt%283%5E2-4%2A1%2A-154%29%29%2F%282%2A1%29
Note: The areas 154 and 460 are included and we deal with positive values because we are looking for measurements.
So we have 11 and 20 from eqn 2 and 1 respectively. However, the possible values for the Width in order to find the areas between 154 and 460 ranges from 11 to 20.
Mathematically: 11%3Cx%3C20 is the solution, 11 and 20 are included.