Maxima, Minima & Optimisation

A critical point (or stationary point) of $f$ is a value $x = c$ in the domain where $f'(c) = 0$ or $f'(c)$ does not exist.

• Local maxima and minima can occur only at critical points — but a critical point need NOT be an extremum (it may be a point of inflection, e.g. $x = 0$ for $f(x) = x^3$).
• So the recipe is always: (1) compute $f'(x)$; (2) solve $f'(x) = 0$ (and note where it is undefined); (3) classify each candidate with the first- or second-derivative test.

At a critical point $c$, examine the sign of $f'$ just to the left and just to the right:

• $f'$ changes $+ \to -$ $\Rightarrow$ local maximum at $c$.
• $f'$ changes $- \to +$ $\Rightarrow$ local minimum at $c$.
• $f'$ does not change sign $\Rightarrow$ neither (a point of inflection).

This test always works — even when $f''$ is awkward to compute or when $f''(c) = 0$ leaves the second-derivative test inconclusive. Factor $f'(x)$ into linear/quadratic pieces and read the sign in each interval.

At a critical point $c$ (where $f'(c) = 0$):

• $f''(c) < 0$ $\Rightarrow$ curve concave down $\Rightarrow$ local maximum.
• $f''(c) > 0$ $\Rightarrow$ curve concave up $\Rightarrow$ local minimum.
• $f''(c) = 0$ $\Rightarrow$ inconclusive — fall back to the first-derivative test.

This is usually the fastest test when $f''$ is easy to compute at the critical point.

'$f$ has an extreme value at $x = p$' means $f'(p) = 0$. With two given extreme points you get two equations in the unknown parameters — a routine linear system. The signature MHT-CET template is $y = a\log x + bx^2 + x$ with extrema at $x = -1$ and $x = 2$:

To find the greatest/least value of $f$ on a set $S$ given by a quadratic inequality: 1. Solve the inequality. $x^2 + 30 \le 11x \Leftrightarrow x^2 - 11x + 30 \le 0 \Leftrightarrow (x-5)(x-6) \le 0 \Leftrightarrow x \in [5, 6]$. 2. Find interior critical points of $f$ that lie inside the interval (often there are none — $f$ may be monotonic on such a short interval). 3. Evaluate $f$ at every critical point in the interval and at both endpoints; the largest is the absolute max, the smallest the absolute min. For $f(x) = 3x^3 - 18x^2 + 27x - 40$, $f'(x) = 9(x-3)^2 \ge 0$, so $f$ is increasing on $[5,6]$ — the max is at $x = 6$: $f(6) = 122$.

The recipe: (1) express the target $Q$ and the constraint; (2) eliminate a variable so $Q = Q(t)$; (3) solve $Q'(t) = 0$; (4) confirm max/min. The AM-GM shortcut: for positive terms, AM $\ge$ GM with equality when the terms are equal. So a sum with fixed product is minimised, and a product with fixed sum is maximised, when the terms are equal:

For an open tank with a square base of side $x$ and height $h$, volume $V = x^2 h$:

• Surface area (base + 4 sides, no top) $A = x^2 + 4xh$.
• Eliminate $h = \dfrac{V}{x^2}$: $A(x) = x^2 + \dfrac{4V}{x}$.
• $A'(x) = 2x - \dfrac{4V}{x^2} = 0 \Rightarrow x^3 = 2V \Rightarrow x = 2h$ (the optimal side is twice the height).

For a cost version, weight each face by its unit cost before minimising. Always confirm with $A'' > 0$ that it is a minimum.

Profit $P(x) = R(x) - C(x)$, where $R(x) = (\text{price per item}) \times x$ is revenue and $C(x)$ is total cost.

• If the price per item is $p(x)$, then $R(x) = x\,p(x)$.
• Maximise: solve $P'(x) = 0$ (marginal revenue = marginal cost) and check $P''(x) < 0$.
• The final answer is the profit VALUE $P(x^*)$ at the optimal $x^*$, unless the number of items itself is asked.

Trig minimum: for $a > b > 0$, the minimum of $a\sec\theta - b\tan\theta$ on $(0, \tfrac{\pi}{2})$ is $\sqrt{a^2 - b^2}$, reached when $\sin\theta = \dfrac{b}{a}$. Harmonic ($a\sin\theta + b\cos\theta$): its extreme values are $\pm\sqrt{a^2 + b^2}$. Rational $\dfrac{x^2 - x + 1}{x^2 + x + 1}$: $y' = \dfrac{2(x^2 - 1)}{(x^2 + x + 1)^2} = 0$ at $x = \pm 1$; $f(-1) = 3$ (max), $f(1) = \tfrac{1}{3}$ (min), so the range is $\big[\tfrac13, 3\big]$.