Perimeter constraint
The perimeter stays fixed, so every width choice forces one matching height.
Rectangle width 3.4 m
Concept module
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.
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. With perimeter 24 m, the rectangle is 3.4 m by 8.6 m for area 29.24 m^2. The local area slope is 5.2. The area curve is still rising here, so making the rectangle a little wider would increase the area. The rectangle is narrower than the best-area square, so it still has room to gain area by trading a little height for width.
Interactive lab
Optimization under constraints
Drag the top-right corner or the objective point to change the width.
Controls
Move the width while the fixed 24 meter perimeter forces the matching height.
Presets
Predict -> manipulate -> observe
Keep the active prompt next to the controls so each change has an immediate visible consequence.
ObservationPrompt 1 of 1
Here the rectangle is still too narrow. The objective curve is rising, so giving up a little height still buys more area.
Graphs
Switch graph views without breaking the live stage and time link.
Objective graph
The constrained area function. The peak marks the best possible rectangle under the fixed perimeter.
width w (m): 0.6 to 11.4area A (m^2): 0 to 40
Area A(w)
Equation map
See each variable before you move it.
Select a symbol to highlight the matching control and the graph or overlay it most directly changes.
Rectangle width
3.4 mMoves the current rectangle width while the fixed perimeter forces the matching height, the objective point, and the local area slope to update together.
Graph: Objective graphGraph: Local area slopeGraph: Constraint tradeoffOverlay: Constraint bandOverlay: Best rectangle overlayOverlay: Area guides
Equations in play
Choose an equation to sync the active symbol, control highlight, and related graph mapping.
More tools
Detailed noticing prompts, guided overlays, and challenge tasks stay available without taking over the main bench.
Hide
More tools
Detailed noticing prompts, guided overlays, and challenge tasks stay available without taking over the main bench.
What to notice
Use the rectangle, the objective graph, and the slope graph as one bench.
ObservationPrompt 1 of 1
Graph: Objective graph
Here the rectangle is still too narrow. The objective curve is rising, so giving up a little height still buys more area.
Control: Rectangle widthGraph: Objective graphGraph: Local area slopeOverlay: Constraint bandOverlay: Area guides
Guided overlays
Focus one overlay at a time to see what it represents and what to notice in the live motion.
3 visible
Overlay focus
Constraint band
Keep the fixed perimeter visible on the current rectangle.
What to notice
- Every time width grows, height must shrink because the perimeter is staying fixed at 24 meters.
Why it matters
It keeps the optimization story honest by showing the tradeoff instead of hiding it behind symbols.
Control: Rectangle widthGraph: Constraint tradeoffEquation
Challenge mode
Use the rectangle, the objective graph, and the slope readout together. The goal is to land the maximum for a reason, not just to hunt 36.
0/1 solved
MaximizeCore
4 of 5 checks
Find the square maximum
Move the width until the fixed-perimeter rectangle reaches the maximum area and the local area slope is essentially zero.
Graph-linkedGuided start2 hints
Suggested start
Start from the narrow rectangle, then move toward the square while watching the height tradeoff and the objective turn flat.
Matched
Open the Objective graph graph.
Objective graphMatched
Keep the Best rectangle overlay visible.
OnMatched
Keep the Constraint band visible.
OnMatched
Keep the Area guides visible.
OnPending
Bring the width into the square band between and meters.
3.4 mThe checklist updates from the live simulation state, active graph, overlays, inspect time, and compare setup.
With perimeter 24 m, the rectangle is 3.4 m by 8.6 m for area 29.24 m^2. The local area slope is 5.2. The area curve is still rising here, so making the rectangle a little wider would increase the area. The rectangle is narrower than the best-area square, so it still has room to gain area by trading a little height for width.
Equation detailsDeeper interpretation, notes, and worked variable context.
Objective function
Substitutes the fixed-perimeter constraint into the area formula so the whole tradeoff depends on one variable.
Rectangle width 3.4 m
Local area slope
Tells whether making the rectangle a little wider would increase area, decrease area, or land at the turning point.
Rectangle width 3.4 m
Progress
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Stable links
Short explanation
What the system is doing
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.
Worked example
Read the full frozen walkthrough.
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.
Live worked examples are available on Premium. You can still read the full frozen walkthrough on the free tier.
View plansFrozen 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.
Prediction prompt
Decide whether you need to move toward or away from the square before you test it.
Check your reasoning
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?
Choose one answer to reveal feedback, then test the idea in the live system if a guided example is available.
Accessible description
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.
Keep the calculus branch connected
Keep moving through the concept map.
These suggestions come from the concept registry, so the reason label reflects either curated guidance or the fallback progression logic.
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