Find Garden Length & Width: Perimeter & Area System

What This Problem Teaches

• Setting up systems of equations from geometric constraints
• Working with perimeter and area formulas simultaneously
• Solving quadratic equations that arise from real-world contexts
• Interpreting multiple solutions in geometric problems
• Connecting algebra to spatial reasoning

Let's Draw It

The fence forms the perimeter: 2l + 2w = 190 ft
The garden has area: l × w = 2100 ft²

Solution: Method 1 — System of Equations Approach

Let's define our variables and translate the geometric constraints into algebraic equations.

Step 1 — Define variables and set up equations

Let l = length of the garden (in feet) and w = width of the garden (in feet).

From the problem constraints:

Step 2 — Simplify the perimeter equation

Divide the perimeter equation by 2:

Step 3 — Solve for one variable

From l + w = 95, we can express the width in terms of length:

Step 4 — Substitute into the area equation

Replace w with (95 - l) in the area equation:

Step 5 — Rearrange into standard quadratic form

Move all terms to one side:

Step 6 — Solve the quadratic equation

Using the quadratic formula where a = 1, b = -95, c = 2100:

This gives us two solutions:

Step 7 — Find the corresponding widths

For each length value, calculate w = 95 - l:

Solution: Method 2 — Roots and Coefficients Approach

Instead of substitution, we can recognize that length and width are the roots of a quadratic equation.

Step 1 — Identify sum and product

From our constraints, we know:

Step 2 — Construct the quadratic

If two numbers have sum S and product P, they are roots of t² - St + P = 0:

Step 3 — Factor the quadratic

We need two numbers that multiply to 2100 and add to 95. Let's find factors of 2100:

Since 60 + 35 = 95, we have:

Step 4 — Identify the solutions

The roots are t = 60 and t = 35, giving us the same dimensions: 60 ft and 35 ft.

Verification

Let's confirm our answer satisfies both original constraints.

Watch Out For These

The Pattern Behind This

This problem exemplifies a fundamental pattern in optimization and constraint problems.

General Setup: Given a rectangle with perimeter P and area A, the dimensions are roots of:

The discriminant (P/2)² - 4A tells us whether real solutions exist. For our problem:

Since the discriminant is positive and a perfect square, we get two distinct rational solutions. If the discriminant were zero, we'd have a square. If negative, no real rectangle could satisfy both constraints.

Connection to other fields: This same mathematical structure appears in economics (cost-revenue optimization), physics (projectile motion), and engineering (design constraints).

Real Applications

• Landscaping: Designing gardens, pools, or lawns with fixed budgets for edging or fencing materials
• Construction: Optimizing room layouts when both floor area and wall perimeter (for baseboards, crown molding) matter
• Agriculture: Planning crop fields where both planting area and fencing costs must be considered

What If?