The optimisation method

In an optimisation problem one seeks the maximum or minimum of a geometric or physical quantity, which is expressed as a function of one variable (or of a parameter). The method is: (1) write the objective function; (2) differentiate; (3) set the derivative equal to zero; (4) check that it is indeed a maximum or a minimum (with $f''$ or with the sign table).

Example — Minimum hypotenuse

A right-angled triangle has a leg of height $h=8$. Let $x$ be the projection of this leg onto the hypotenuse. Find $x$ that minimises the length of the hypotenuse $i$.

The leg $CH=h$ is the altitude relative to the hypotenuse; $x$ is the projection of one leg, $i-x$ that of the other.

Solution

Setting up. From the second Euclid theorem: $x\cdot(i-x) = h^2 = 64$. Hence: $i = x + \frac{64}{x}, \qquad x > 0.$ Note: if a fraction has several terms in the numerator, before differentiating it is convenient to turn it into a sum of fractions. Here it is already done: $i = x + 64/x = x + 64\cdot x^{-1}$.

Derivative: $i'(x) = 1 - \dfrac{64}{x^2}$.

$i'(x) = 0 \iff x^2 = 64 \iff x = 8$ (we discard $x=-8$ because $x>0$).

Check: $i''(x) = \dfrac{128}{x^3} > 0$ for $x>0$: the function is convex, so $x=8$ is a minimum.

$i_{\min} = 8 + 64/8 = 8 + 8 = \boxed{16}$.

Links

Topics: Derivatives
Concepts: Objective function · Maximum and minimum · Stationary point
Methods: Maximum minimum via derivative
Skills: Modelling · Studying a function