Math Problem Statement

An animal shelter director is planning to build a rectangular playpen. The playpen must have a perimeter of 150 feet and an area of at least 1000 square feet. Describe the possible lengths of the playpen.

Solution

Let's solve the problem step by step.

Problem Recap:

  • A rectangular playpen must have:
    • Perimeter = 150 feet
    • Area ≥ 1000 square feet
  • We are asked to describe the possible lengths of the playpen.

1. Perimeter Equation:

The perimeter PP of a rectangle is given by: P=2L+2WP = 2L + 2W where LL is the length, and WW is the width.

For this problem: 2L+2W=1502L + 2W = 150 We can simplify: L+W=75(Equation 1)L + W = 75 \quad \text{(Equation 1)} So: W=75L(solve for W in terms of L)W = 75 - L \quad \text{(solve for } W \text{ in terms of } L)


2. Area Constraint:

The area AA of the rectangle is: A=L×WA = L \times W Using the expression for WW from Equation 1: A=L×(75L)1000A = L \times (75 - L) \geq 1000 Expanding: A=75LL21000A = 75L - L^2 \geq 1000 Rearrange: L275L+10000L^2 - 75L + 1000 \leq 0


3. Solving the Quadratic Inequality:

First, solve the corresponding equation: L275L+1000=0L^2 - 75L + 1000 = 0 Using the quadratic formula: L=(75)±(75)24(1)(1000)2(1)L = \frac{-(-75) \pm \sqrt{(-75)^2 - 4(1)(1000)}}{2(1)} L=75±562540002L = \frac{75 \pm \sqrt{5625 - 4000}}{2} L=75±16252L = \frac{75 \pm \sqrt{1625}}{2} 162540.31\sqrt{1625} \approx 40.31 L=75±40.312L = \frac{75 \pm 40.31}{2} So: L=75+40.31257.66andL=7540.31217.34L = \frac{75 + 40.31}{2} \approx 57.66 \quad \text{and} \quad L = \frac{75 - 40.31}{2} \approx 17.34


4. Length Range:

The solutions to the quadratic equation are L17.34L \approx 17.34 and L57.66L \approx 57.66. Since the quadratic opens upward (positive L2L^2 coefficient), the inequality L275L+10000L^2 - 75L + 1000 \leq 0 holds between these two values.

Thus, the possible lengths are: 17.34L57.6617.34 \leq L \leq 57.66


Final Answer:

The length of the playpen must be at least 17.34 feet and at most 57.66 feet.


Do you want further clarification on any step or need more examples?

Here are 5 follow-up questions to deepen your understanding:

  1. How would the dimensions change if the perimeter was 160 feet instead of 150?
  2. What happens if the required area is increased to 1200 square feet?
  3. Can you determine the exact width for the minimum and maximum lengths found?
  4. How does changing only the width affect the area and perimeter?
  5. Can you verify the answer by substituting the limits back into the perimeter and area formulas?

Tip: Always try to sketch a quick graph when working with quadratic inequalities to visualize the solution.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Inequalities
Geometry

Formulas

Perimeter of a rectangle: P = 2L + 2W
Area of a rectangle: A = L * W
Quadratic equation: ax^2 + bx + c = 0

Theorems

Quadratic formula: L = [-b ± sqrt(b^2 - 4ac)] / 2a

Suitable Grade Level

Grades 9-11