Optimisation

Steps

Draw the situation if possible.
Find the target function, i.e. the function to optimise.
Find the constraint equation linking the variables.
Rewrite the target function in terms of a single variable.
Find the domain of the target function.
Differentiate the target function to find the critical points in the domain.
Identify the extremum by using:
- First / second derivative test
- Other arguments, e.g. vertex of a parabola

The $\frac{\mathrm{d}}{\mathrm{d}x}f(x) = f'(x)$ step can be omitted for optimisation.

Examples

Two poles, one 15 meters tall and one 10 meters tall, are 40 meters apart.
A length of wire is attached to the top of each pole, and it is staked to the
ground somewhere between the two poles. Where should the wire be
staked so that the angle formed by the two pieces of wire at the stake is a
maximum?

Optimisation 1

Note that $θ = π - \arctan \frac{10}{x} - \arctan \frac{15}{40 - x}$ .
So,

\begin{aligned} θ^{'} & = - (\frac{- \frac{10}{x^{2}}}{1 + {(\frac{10}{x})}^{2}} + \frac{\frac{15}{(40 - x)^{2}}}{1 + {(\frac{15}{40 - x})}^{2}}) \\ = - (\frac{- 10}{x^{2} + 10^{2}} + \frac{15}{(40 - x)^{2} + 15^{2}}) \\ = - (- \frac{- 10 ((40 - x)^{2} + 15^{2}) + 15 (x^{2} + 10^{2})}{(x^{2} + 10^{2}) ((40 - x)^{2} + 15^{2})}) \\ = \frac{- 5 (x^{2} + 160 x - 3350)}{(x^{2} + 10^{2}) ((40 - x)^{2} + 15^{2})} \end{aligned}

$θ$ is never undefined. $θ^{'} = 0$ if

\begin{array}{r} x^{2} + 160 x - 3350 = 0 \\ (x + 80)^{2} - 9750 = 0 \\ (x + 178.74 \dots) (x - 18.74 \dots) = 0 \\ ⟹ x = - 178.74 \dots or 18.74 \dots \end{array}

We reject $x_{-}$ because $x > 0$ .

0	$x_{+}$	40
$θ^{'}$	+	-
$θ$	inc	dec

By the first derivative test, $θ$ has a global maximum at $x = x_{+}$ .
Thus, the wires should be stacked at about 18.74m from the 10m pole.

(The angle is:)

\begin{aligned} θ (- 80 + \sqrt{80^{2} - 3350}) & = π - \arctan \frac{10}{- 80 + \sqrt{80^{2} - 3350}} - \arctan \frac{15}{40 + 80 - \sqrt{80^{2} - 3350}} \\ \approx 3.30 \dots rad \end{aligned}

A box (with no top) is to be constructed from a piece of cardboard of sides
15ft and 24ft by cutting out squares of length ℎ from the corners and folding
up the sides. Find the value of ℎ that maximizes the volume of the box. What
are the dimensions of this box?

We need $15 - 2 h > 0$ and $24 - 2 h > 0$ . The former implies that $h < 7.5$ while the latter says $h < 12$ . So, $h \in (0, 7.5)$ .
Then,

\begin{aligned} V (h) & = (15 - 2 h) (24 - 2 h) h \\ = (15 \cdot 24 - 30 h - 48 h + 4 h^{2}) h \\ = 360 h - 78 h^{2} + 4 h^{3} \end{aligned}

So,

\begin{aligned} V^{'} (h) & = 360 - 156 h + 12 h^{2} \\ V^{″} (h) & = - 156 + 24 h \end{aligned}

$V^{'}$ is never undefined.

\begin{aligned} V^{'} (h) & = 0 \\ 360 - 156 h + 12 h^{2} & = 0 \\ h & = \frac{156 \pm \sqrt{(- 156)^{2} - 4 \cdot 12 \cdot 360}}{2 \cdot 12} \\ h & = 10 or 3 \end{aligned}

Since $h \in (0, 7.5)$ , we reject $h = 10$ .
Then,

\begin{aligned} V^{″} (3) & = - 156 + 24 \cdot 3 \\ = - 84 \end{aligned}

By the second derivative test, V has a global maximum at $x = 3$ .
Finally,

\begin{aligned} 15 - 2 h & = 9 ft \\ 24 - 2 h & = 18 ft \end{aligned}

The dimensions of the box are 18x9x3ft.

Contributors

Ajitani Hifumi

Changelog

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Optimisation ​

Steps ​

Examples ​

Contributors

Changelog

Optimisation

Steps

Examples