Question

the perimeter of a rectangle is to be no greater than 80 centimeters and the length must be 25 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 80. x + 25 + x + 25 ≤ 80. simplify the left - side of the inequality. x + 25 + x + 25 ≤ 80. 2x + 50 ≤ 80 (simplify your answer. do not factor.) apply the addition property of inequality. 2x + 50 ≤ 80. 2x ≤ □ (simplify your answer.)

Explanation:

Step1: Recall perimeter formula

The perimeter $P$ of a rectangle is $P = 2l+2w$, where $l$ is length and $w$ is width. Given $l = 25$ and $P\leq80$, we have $2\times25 + 2x\leq80$ (let width be $x$).

Step2: Simplify left - hand side

$2\times25+2x=50 + 2x$, so the inequality becomes $2x + 50\leq80$.

Step3: Apply subtraction property

Subtract 50 from both sides of the inequality $2x+50\leq80$. We get $2x+50 - 50\leq80 - 50$, which simplifies to $2x\leq30$.

Step4: Solve for $x$

Divide both sides of the inequality $2x\leq30$ by 2. So $x\leq15$.

Answer:

15