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Optimization / Maxima, Minima, and Constraints

Move one rectangle width under a fixed perimeter, watch the area curve peak, and use the local slope to see why the square is the best constrained shape.

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Featured setups

Narrow and rising

Open a narrow rectangle where the objective curve is still climbing.

Square at the peak

Open the best-area square with the peak point and optimum overlay visible.

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Stable links

Why it behaves this way

Explanation

Optimization becomes easier to trust when one constrained picture stays visible. This bench keeps one rectangle under a fixed 24 meter perimeter, lets you move only the width, and forces the height, the area, and the objective graph to respond together so maxima and minima stay tied to a real tradeoff instead of a detached worksheet trick.

The important idea is that width and height are not free to change independently. Every extra meter of width costs one meter of height because the perimeter is fixed. The area curve peaks when that tradeoff has stopped helping, and the local slope of the objective graph is the cleanest way to see that turning point.

Key ideas

01A constraint turns one quantity into a tradeoff with another, so optimization is about the whole relationship, not about making one side as large as possible.
02For a fixed 24 meter perimeter, the rectangle area can be written as one objective function A(w) = w(12 - w).
03The maximum area occurs where the objective graph flattens and changes from rising to falling, which is why A'(w) = 0 is the key local-rate test.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough
Use the current rectangle state and the peak square from the same fixed-perimeter bench. These walkthroughs keep the algebra tied to the live geometry and graphs instead of splitting the optimization story into a second worksheet.

Premium unlocks saved study tools, exact-state sharing, and the richer review surfaces that support this guided flow.

View plans
Frozen valuesUsing frozen parameters

For the current width of meters, what height and area does the fixed 24 meter perimeter force?

Rectangle width

3.4 m

Rectangle height

8.6 m

1. Rewrite the perimeter constraint

With , the fixed half-perimeter is .

2. Solve for the matching height

At , the matching height is .

3. Compute the constrained area

So .

Current constrained area

This rectangle is still 6.76 square meters below the best case, and making it wider would move it toward the peak.

Common misconception

The maximum area must happen at the widest rectangle because a larger width always means a larger area.

A wider rectangle also has to become shorter when the perimeter is fixed, so width and height are trading against each other.

Past the best square, the lost height hurts the product more than the extra width helps it, so the area curve turns downward.

Mini challenge

Move the width until the rectangle is still not a square, but the objective graph is almost flat.

Make a prediction before you reveal the next step.

Decide whether you need to move toward or away from the square before you test it.

Check your reasoning against the live bench.

You need to move toward the square, because the area slope only approaches zero as the width and height become equal under this fixed perimeter.
The peak happens where the objective graph changes from rising to falling. That is the local-rate signal that the square is the maximum-area rectangle here.

Quick test

Graph reading

Question 1 of 3

Use the constraint, objective graph, and local slope together. These checks are about why the maximum happens.

At the best-area rectangle, what must the local slope of the area curve be?

Use the live bench to test the result before moving on.

Accessibility

The simulation shows one rectangle under a fixed 24 meter perimeter. Dragging the top-right corner changes the width and automatically adjusts the height. A smaller objective graph shows the matching point on the area curve, and an optional square overlay marks the best-area case.

Moving the width updates the rectangle dimensions, the area readout, and the local slope of the objective together so the learner can compare the constrained shape and the turning point on one compact bench.

Graph summary

The objective graph plots area against width for the fixed perimeter, so its peak marks the maximum-area rectangle. The local-slope graph plots A'(w), which crosses zero at the same width.

The constraint-tradeoff graph shows h = 12 - w, making the one-for-one exchange between width and height visible instead of hidden.