Question

the perimeter of a rectangle is to be no greater than 120 centimeters and the length must be 40 centimeters. find the maximum width of the rectangle. first, understand the problem. then translate the statement into an inequality. the perimeter of the rectangle is less than or equal to 120 x + 40 + 120

Explanation:

Step1: Recall perimeter formula

The perimeter $P$ of a rectangle is $P = 2(l + w)$, where $l$ is length and $w$ is width. Here $l = 40$ and let $w=x$. So $P=2(40 + x)=x + 40+x + 40$.

Step2: Set up the inequality

Since the perimeter is no greater than 120, we have $x + 40+x + 40\leq120$. Simplifying gives $2x+80\leq120$.

Step3: Solve the inequality

Subtract 80 from both sides: $2x+80 - 80\leq120 - 80$, so $2x\leq40$. Then divide both sides by 2: $\frac{2x}{2}\leq\frac{40}{2}$, resulting in $x\leq20$.

Answer:

20 centimeters