Write and solve an inequality to find the possible values of x if the maximum area of the rectangle is to be 63 square meters. rectangle 3 meters high x (2x + 1) meters wide

L • W ≤ Area

3 • x (2x + 1) ≤ 63

3x (2x + 1) ≤ 63

6x^2 + 3x ≤ 63

6x^2 + 3x - 63 ≤ 0

We're going to do AC method to find the x values:

1) Find what 2 numbers equal a (6) and c (-63) multiplied and equal b (3) when added. These two numbers are 21 and - 18

2) Expand the equation into these two numbers

6x^2 - 18x + 21x - 63 ≤ 0

3) Group and factor

6x (x - 3) + 21 (x - 3) ≤ 0

(6x + 21) (x - 3) ≤ 0

4) Set each parentheses section to 0.

6x + 21 ≤ 0 x-3 ≤ 0

6x ≤ - 21 x ≤ 3

x ≤ - 21/6

... or - 3.5

In conclusion, the x values can be x ≤ - 3.5 or x ≤ 3.