Let's analyze the problem step by step.

Problem Summary:

You need to find the possible lengths of a rectangular garden where:

1. The perimeter is 140 feet.
2. The area is at least 1,000 square feet.

Variables:

Let the length of the garden be $l$ feet, and the width be $w$ feet.

Step 1: Perimeter Equation

The perimeter $P$ of a rectangle is given by: $P = 2l + 2w$ In this case, the perimeter is 140 feet, so: $2l + 2w = 140 \quad \text{or} \quad l + w = 70$ This equation simplifies to $l + w = 70$.

Step 2: Area Equation

The area $A$ of the rectangle is given by: $A = l \times w$ We know that the area must be at least 1,000 square feet, so: $l \times w \geq 1000$

Step 3: Express Width in Terms of Length

From the perimeter equation $l + w = 70$, we can express the width in terms of the length: $w = 70 - l$

Step 4: Substitute into the Area Equation

Substitute $w = 70 - l$ into the area equation $l \times w \geq 1000$: $l \times (70 - l) \geq 1000$ This simplifies to: $70l - l^2 \geq 1000$ Rearranging the terms: $l^2 - 70l + 1000 \leq 0$

Step 5: Solve the Quadratic Inequality

To solve $l^2 - 70l + 1000 \leq 0$, we can use the quadratic formula: $l = \frac{-(-70) \pm \sqrt{(-70)^2 - 4(1)(1000)}}{2(1)}$ Simplifying: $l = \frac{70 \pm \sqrt{4900 - 4000}}{2}$ $l = \frac{70 \pm \sqrt{900}}{2}$ $l = \frac{70 \pm 30}{2}$ So the two possible solutions are: $l = \frac{70 + 30}{2} = 50 \quad \text{or} \quad l = \frac{70 - 30}{2} = 20$

Step 6: Determine the Length Range

• If $l = 50$, then $w = 70 - 50 = 20$.
• If $l = 20$, then $w = 70 - 20 = 50$.

Therefore, the approximate length of the garden is at least 20 feet and at most 50 feet.

Let me know if you'd like further clarification or additional details on this! Here are 5 follow-up questions you might explore:

1. How would the solution change if the area had to be exactly 1,000 square feet instead of at least 1,000 square feet?
2. What if the perimeter of the garden were different—say, 150 feet?
3. Can you visualize this garden's possible dimensions on a graph of the quadratic equation $l^2 - 70l + 1000 = 0$?
4. How would you solve for the possible dimensions if the garden were a triangle instead of a rectangle?
5. What if we changed the shape of the garden to a circle but kept the same area and perimeter constraints?

Tip:

For quadratic inequalities, always check the solutions of the corresponding quadratic equation first to determine the intervals where the inequality holds.