Math · Calculus

Optimization / Maxima, Minima, and Constraints

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Wrap-up

What you learned

Recommended next: Open concept testCheck whether the core ideas are ready without leaving this concept.
Read next: Derivative as Slope / Local Rate of ChangeReview the local-slope signal behind the maximum

Key takeaway

A fixed 24 meter perimeter links width and height by h = 12 - w.
The objective A(w) = w(12 - w) is the rectangle area after the constraint has removed the extra variable.
The derivative sign tells which direction improves the area, and A'(w) = 0 marks the square maximum.
The widest rectangle does not win because extra width eventually costs too much height.

Common misconception

The largest width is not the largest area; under a fixed perimeter, width and height trade off, so the product peaks at the balanced square.

On this bench, making the width larger always forces the height smaller because the perimeter stays fixed at 24 meters.

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Integral as Accumulation / Area

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Equation snapshotPerimeter constraint · Objective functionShow 3 equations

Perimeter constraint
$2 w + 2 h = 24$
Fixes the total perimeter, so choosing a width immediately fixes the matching height.
Objective function
$A (w) = w (12 - w)$
Rewrites the rectangle area using the perimeter constraint so the whole optimization problem depends on one variable.
Local area slope
$A^{'} (w) = 12 - 2 w$
Tells whether making the rectangle a little wider would increase area, decrease area, or land at the turning point.

Why it behaves this way

Explanation

Optimization becomes easier to understand when the shape, the constraint, and the objective graph all respond to one choice. On this bench the perimeter is fixed at 24 meters, so choosing the width automatically fixes the height, the rectangle, the area, and the point on the graph.

That is why the square maximum is not a trick to memorize. On the narrow side, gaining width helps more than the lost height hurts. Past the square, the lost height hurts more than the extra width helps. The best rectangle sits at the balance point where the objective curve reaches its peak and the local area slope is zero.

Key ideas

01With perimeter 24 m, width and height are linked by h = 12 - w, so one width choice already determines the whole rectangle.

02The objective graph A(w) = w(12 - w) is the same rectangle problem written in one variable, so its peak is the best possible area under this constraint.

03A'(w) tells you which direction improves the area: positive means making the rectangle wider still helps, negative means you have gone past the best shape, and zero marks the turning point.

Worked examples

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Frozen walkthrough

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Frozen walkthrough

Use the width slider, the rectangle, and the graphs together. First read the forced height from the perimeter, then read the area, then use the slope graph to decide whether wider would help or hurt.

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Example 1 of 2

Frozen valuesUsing frozen parameters

At the current width of $3.2$ meters, what height does the fixed perimeter force, and what area does that give?

w

Rectangle width

3.2 m

h

Rectangle height

8.8 m

1. Use the fixed-perimeter constraint

With

2 w + 2 h = 24

, the fixed half-perimeter is

w + h = 12

2. Find the matching height

w = 3.2

, the matching height is

h = 12 - 3.2 = 8.8

3. Compute the area from width times height

A = w h = 3.2 (8.8) = 28.16

Current area under the constraint

A = 28.16 t e x t m^{2}

This rectangle is still 7.84 square meters below the best case, and making it wider would move it toward the peak.

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