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.