Math Problem Statement

A new building is being constructed at UT Austin. It is 200 feet tall, and has a rectangular 100 ft by 150 ft footprint. The building has a walkway that is x feet wide surrounding all four sides of the building. If the total area (around and including the building) is 16,536 square feet, how wide is the walkway? Note: Your answer should be an integer.

Solution

To solve this problem, let's break it down into steps.

  1. Identify the dimensions of the building and the walkway:

    • The building has a rectangular footprint of 100 ft×150 ft100 \text{ ft} \times 150 \text{ ft}.
    • There is a walkway of width xx feet surrounding the building on all four sides.
  2. Calculate the area of the building alone:

    • The area of the building is: 100×150=15,000 square feet100 \times 150 = 15,000 \text{ square feet}
  3. Set up the equation for the total area including the walkway:

    • With the walkway, the dimensions of the entire area become:
      • Width: 100+2x100 + 2x
      • Length: 150+2x150 + 2x
    • The total area, including the walkway, is given to be 16,536 square feet. Therefore, we can set up the equation: (100+2x)(150+2x)=16,536(100 + 2x)(150 + 2x) = 16,536
  4. Expand and solve the equation:

    • Expanding the left side: 100×150+100×2x+150×2x+2x×2x=16,536100 \times 150 + 100 \times 2x + 150 \times 2x + 2x \times 2x = 16,536 15,000+200x+300x+4x2=16,53615,000 + 200x + 300x + 4x^2 = 16,536 4x2+500x+15,000=16,5364x^2 + 500x + 15,000 = 16,536
    • Simplify the equation by moving 16,536 to the left side: 4x2+500x+15,00016,536=04x^2 + 500x + 15,000 - 16,536 = 0 4x2+500x1,536=04x^2 + 500x - 1,536 = 0
    • Divide the equation by 4 for simplicity: x2+125x384=0x^2 + 125x - 384 = 0
  5. Solve the quadratic equation x2+125x384=0x^2 + 125x - 384 = 0:

    • Using the quadratic formula: x=125±1252+4×3842×1x = \frac{-125 \pm \sqrt{125^2 + 4 \times 384}}{2 \times 1} x=125±15625+15362x = \frac{-125 \pm \sqrt{15625 + 1536}}{2} x=125±171612x = \frac{-125 \pm \sqrt{17161}}{2} x=125±1312x = \frac{-125 \pm 131}{2}
    • Solving this, we get two values for xx: x=62=3orx=2562=128x = \frac{6}{2} = 3 \quad \text{or} \quad x = \frac{-256}{2} = -128
    • Since xx represents a width, it must be positive, so x=3x = 3.

Conclusion:

The width of the walkway is 3 feet\mathbf{3 \text{ feet}}.

Would you like more details or have any questions? Here are some related questions:

  1. How would the width of the walkway change if the building had a different footprint?
  2. What if the total area were different, say 18,000 square feet? How would we solve for xx?
  3. Can you derive a general formula for the width of the walkway based on the building dimensions and total area?
  4. What happens to the dimensions of the walkway if we have an irregular-shaped building?
  5. How would this problem change if there were a circular building instead of a rectangular one?

Tip: Always check if your answer is reasonable by plugging it back into the original problem to verify it meets all given conditions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometry
Quadratic Equations

Formulas

Area of a rectangle: width * length
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic formula

Suitable Grade Level

Grades 9-12