Mathematics

Calculus: Optimization

Derivatives, critical points, and applied maxima — the workhorse of calculus.

To optimize $f (x)$ , find critical points where $f^{'} (x) = 0$ . The second-derivative test classifies them: $f^{''} (x) < 0$ marks a local maximum, $f^{''} (x) > 0$ a local minimum.

1. A single maximum
easy
Find and classify the critical point of $f (x) = - 2 x^{2} + 8 x + 3$ .
Show solution
$f^{'} (x) = - 4 x + 8 = 0 \Rightarrow x = 2$ .

$f^{''} (x) = - 4 < 0$ , so $x = 2$ is a local maximum.

$f (2) = - 2 (4) + 16 + 3 = 11$ . The maximum value is $11$ at $x = 2$ .
2. Two critical points
medium
Locate and classify all critical points of $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ .
Show solution
$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3) = 0 \Rightarrow x = 1, 3$ .

$f^{''} (x) = 6 x - 12$ . At $x = 1$ : $f^{''} (1) = - 6 < 0$ → local maximum, $f (1) = 5$ . At $x = 3$ : $f^{''} (3) = 6 > 0$ → local minimum, $f (3) = 1$ .
3. Revenue maximization
hard
A firm faces demand $p = 100 - 2 q$ . Revenue is $R (q) = pq$ . Find the quantity that maximizes revenue and the maximum revenue.
Show solution
$R (q) = (100 - 2 q) q = 100 q - 2 q^{2}$ .

$R^{'} (q) = 100 - 4 q = 0 \Rightarrow q = 25$ .

$R^{''} (q) = - 4 < 0$ , confirming a maximum. Price $p = 100 - 2 (25) = 50$ ; maximum revenue $R = 50 \times 25 = \$ 1{,}250$.

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Calculus: Optimization

1. A single maximum

2. Two critical points

3. Revenue maximization

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