Math Problem Statement

The perimeter of a rectangle is to be no greater than 130 centimeters and the width must be 30 centimeters. Find the maximum length of the rectangle.

Solution

To solve this problem, let's recall the formula for the perimeter PP of a rectangle:

P=2(length+width)P = 2(\text{length} + \text{width})

We are given:

  • The perimeter should be no greater than 130 cm: P130P \leq 130
  • The width is 30 cm: width=30\text{width} = 30

Now, substitute these values into the perimeter formula:

1302(length+30)130 \geq 2(\text{length} + 30)

Next, solve for the length:

  1. Divide both sides by 2: 65length+3065 \geq \text{length} + 30

  2. Subtract 30 from both sides: 35length35 \geq \text{length}

Thus, the maximum possible length of the rectangle is 35 cm.

Conclusion:

The maximum length of the rectangle is 35 centimeters.

Would you like more details on this problem, or do you have any questions?


Here are 5 related questions:

  1. What would the maximum length be if the perimeter were reduced to 120 cm?
  2. How would you find the area of the rectangle with the maximum length?
  3. What if both width and length are variables—how would you optimize the dimensions?
  4. If the width were increased to 40 cm, what would the new maximum length be?
  5. How does changing the shape to a square affect the perimeter?

Tip: When solving perimeter problems, always start by writing out the perimeter formula and substituting known values.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Geometry

Formulas

Perimeter of a rectangle: P = 2(length + width)

Theorems

Linear equation solving

Suitable Grade Level

Grades 6-8