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 ≤ 80 (simplify your answer. do not factor.)

Explanation:

Step1: Combine like - terms

Combine the \(x\) terms and the constant terms on the left - hand side of the inequality \(x + 25+x + 25\leq80\). We get \((x+x)+(25 + 25)\leq80\), which simplifies to \(2x+50\leq80\).

Step2: Isolate the variable term

Subtract 50 from both sides of the inequality \(2x+50\leq80\). Using the subtraction property of inequalities, we have \(2x+50 - 50\leq80 - 50\), which gives \(2x\leq30\).

Step3: Solve for \(x\)

Divide both sides of the inequality \(2x\leq30\) by 2. By the division property of inequalities, \(\frac{2x}{2}\leq\frac{30}{2}\), so \(x\leq15\).

Answer:

The maximum width of the rectangle is 15 centimeters.