Optimization: Maximum Area with Constraint

Let's Picture This

Before diving into equations, let's visualize what we're working with. This diagram shows our rectangular plot against the river:

Notice that we need fencing for three sides: both sides perpendicular to the river (each of length y) and one side parallel to the river (length x). The river provides the fourth "wall" for free.

What You Will Learn

• How to set up constraint optimization problems using calculus
• Converting a two-variable problem into a single-variable function
• Finding maximum values using derivatives and the first derivative test
• Why geometric optimization problems often have elegant, symmetric solutions
• How to verify optimization results using the second derivative test

Solution: Method 1 — Single-Variable Calculus Approach

Step 1 — Define variables and set up the constraint

Let x = width of the plot (parallel to the river) and y = depth of the plot (perpendicular to the river).

Since one side is along the river, we only need fencing for three sides:

Step 2 — Express area as a function of one variable

The area of the rectangle is A = xy. From our constraint, we can solve for x:

Substituting this into the area formula:

Step 3 — Find the maximum using calculus

To find the maximum area, we take the derivative and set it equal to zero:

Step 4 — Find the corresponding x-value

Using our constraint equation:

Step 5 — Verify this is indeed a maximum

We check the second derivative:

Since the second derivative is negative, we have a maximum.

Solution: Method 2 — Completing the Square

Step 1 — Start with the area function

From Method 1, we have A(y) = 1120y - 2y².

Step 2 — Factor and complete the square

Factor out the coefficient of y²:

Complete the square inside the parentheses:

Step 3 — Rewrite in vertex form

Step 4 — Identify the maximum

This is in vertex form A(y) = a(y - h)² + k where a = -2, h = 280, and k = 156,800.

Since a < 0, the parabola opens downward, so the vertex represents the maximum. The maximum occurs at y = 280 meters, giving x = 560 meters.

Verification

Let's verify our answer satisfies all conditions:

Fence constraint check:

Area calculation:

Maximum verification: Let's test nearby values. If y = 275, then x = 1120 - 2(275) = 570, giving area 275 × 570 = 156,750 m². If y = 285, then x = 1120 - 2(285) = 550, giving area 285 × 550 = 156,750 m². Both are smaller than our answer of 156,800 m², confirming we found the maximum.

Does This Seem Reasonable?

Our answer tells us the optimal plot is 560m × 280m. Notice something interesting: the width along the river is exactly twice the depth. This isn't a coincidence!

We can also check this makes intuitive sense by considering extreme cases:

• If the plot were very narrow (x small), we'd waste fencing on long sides perpendicular to the river
• If the plot were very shallow (y small), we'd waste fencing on the long side parallel to the river
• Our solution balances these competing demands optimally

Watch Out For These

The Pattern Behind This

This problem represents a classic constrained optimization situation. Here's the general approach that works for any similar problem:

The river fence pattern specifically: For any rectangular enclosure where one side needs no fencing, the optimal dimensions satisfy:

This pattern appears in many contexts: fencing against walls, buildings, cliffs, or any natural boundary that provides one "free" side.

You'll See This Again In...

• Engineering: Designing efficient storage tanks, warehouses, and enclosures where one side is against an existing structure
• Agriculture: Planning pastures and crop fields where natural boundaries (rivers, hills, existing fences) provide free containment
• Urban planning: Optimizing park layouts, parking lots, and building footprints against property lines
• Economics: Similar optimization appears in production theory when one resource is "free" or unlimited

What Comes Next

This problem is your entry point to a rich world of optimization. More advanced versions include:

Multiple constraints: What if you also had a maximum depth restriction due to soil conditions?

Non-uniform costs: What if fencing parallel to the river costs differently than perpendicular fencing?

Non-rectangular shapes: Advanced calculus courses explore optimizing circular and elliptical enclosures using Lagrange multipliers.

The systematic thinking you've learned here—expressing everything in terms of one variable, then optimizing—will serve you throughout calculus and beyond.

What If?

Practice your optimization skills with these variations: