Extrema and Optimization

Once you can differentiate, the most natural question appears: where is this function highest, and where lowest? A fence enclosing the largest area, a price earning the largest profit — both are hunts for a "peak". Calculus turns peak-hunting into a procedure: differentiate, set to zero, read the signs. Three steps, and peaks and valleys reveal themselves.

Critical points: where the derivative is zero

The instant you stand on a summit, you are going neither up nor down — the slope is zero. Peaks and valleys of a function are the same: the tangent is horizontal and the derivative vanishes. Solutions of $f'(x) = 0$ are called critical points. Note that this is only a shortlist: every peak and valley appears on it, but not everyone on it is a peak or valley (some are just a flat stretch). A function can carry several critical points — like a winding road with several summits and hollows — and the procedure runs once for each candidate. So step one is always to solve $f'(x) = 0$ and summon the candidates.

Reading peaks and valleys from signs

The sign of the derivative on either side of a critical point announces its identity:

Positive then negative: up, then down — a local maximum (a summit);
Negative then positive: down, then up — a local minimum (a valley bottom);
Same sign on both sides: a level crossing, neither peak nor valley.

InteractiveSlope-Flip Observatory

Point x₀0.5Step h1.5

Secant slope

2.500

Tangent slope

1.000

Drag h toward 0 and the secant hugs the tangent — that's the derivative.

Drag $x_0$ toward 0 and watch this parabola's slope flip from negative to positive, passing exactly through zero at the valley bottom — negative then positive, a textbook local minimum. Every function's peaks and valleys obey the same rule of sign flips.

Two worked problems

Two problems — one geometric, one commercial — run the exact same procedure.

Largest area from 20 meters of fence

Fence a rectangle with 20 meters. Let the width be $x$ ; the length is then $10 - x$ , and the area is

$A(x) = x(10 - x) = 10x - x^2$

Differentiate: $A'(x) = 10 - 2x$ . Setting it to zero gives $x = 5$ . Check the signs: at $x = 3$ , $A' = 4 > 0$ (still climbing); at $x = 7$ , $A' = -4 < 0$ (already falling) — positive then negative, a genuine summit. The length is 5 as well, so the largest area is $5 \times 5 = 25$ square meters: the square wins.

Pricing for maximum profit

Each item costs 10 to make, and at a price of $p$ you sell $(100 - p)$ items. The profit is

$P(p) = (p - 10)(100 - p) = -p^2 + 110p - 1000$

Differentiate: $P'(p) = -2p + 110$ , zero at $p = 55$ . Check the signs: at $p = 50$ , $P' = 10 > 0$ ; at $p = 60$ , $P' = -10 < 0$ — another positive-then-negative summit. Price at 55, sell 45 items, and the profit is $45 \times 45 = 2025$ .

Do not forget the endpoints

On a limited domain, the extreme values may hide at the ends — where the slope exists on one side only, the derivative need not be zero. Real domains usually arrive with edges attached: a fence width cannot be negative, and a price cannot fall below cost — write the boundaries down first, or the endpoint check has nothing to aim at. Both problems above got the courtesy check: the fence endpoints $x = 0$ and $x = 10$ give area 0, and the pricing endpoints $p = 10$ (working for nothing) and $p = 100$ (no buyers) give profit 0 — far below the interior champions 25 and 2025. The full procedure is three steps plus one look: solve $f'(x) = 0$ , check the signs, compare the endpoints. Peaks and valleys of parabolas are easy to spot on a graph — see Graphing Quadratics; the differentiation procedure comes from Introduction to Derivatives, with a quick reference in Derivative Rules; the next stop downstream is Integration.

Check yourself

Quick quiz

1. Left of a point f′ is positive, right of it negative. What is this point?
2. The derivative of A(x) = 10x − x² is?
3. Is a critical point guaranteed to be a maximum or minimum?