Classical Optimisation

Problem 1. Let $f$ be a non-negative function satisfying the following conditions:

$f$ is continuous on $\mathbb R$ ,
$f$ is differentiable on $\mathbb R \backslash \{0\}$ ,
$f$ is strictly increasing on $(-\infty, 0)$ ,
$f$ is strictly decreasing on $(0, \infty)$ , and
$f(|x|) \to 0$ as $x \to \infty$ .

Let $A(x)$ denote the rectangle under the curve $y=f(x)$ whose base that intersects the positive $x$ -axis has length $x$ . If $A$ has a minimum prove that $A$ is minimised when

$\begin{aligned} \left( 1 - \frac { f'(x) }{(g' \circ g^{-1} \circ f)(x)} \right) \cdot f(x) + (x - (g^{-1} \circ f)(x)) \cdot f'(x) &= 0, \end{aligned}$

where $g = f|_{(-\infty, 0)}$ .

(Click for Solution)

Solution. Fix $c > 0$ . If $f(c) = 0$ , then $f$ being strictly decreasing implies $f(c+1) < 0$ , a contradiction. Therefore, $f(0) > f(c) > 0$ . Since $f(-x) \to 0$ as $x \to \infty$ , there exists $b \in (-\infty, 0)$ such that $f(b) < f(c) < f(0)$ . Use the intermediate value theorem to obtain $a \in (b, 0)$ such that $f(a) = f(c)$ . It is not hard to verify that $f(x) > f(c)$ for $x \in [a, c]$ . Therefore, the rectangle has an area $A(c)$ of $A(c) = (c - a) \cdot f(c)$ .

In fact, more is true. Since $g := f|_{(-\infty, 0)}$ is differentiable and strictly decreasing,

$g(a) = f(a) = f(c)\quad \Rightarrow \quad a = g^{-1}(f(c)).$

Hence, $A(x) = (x - (g^{-1} \circ f)(x) ) \cdot f(x)$ . Differentiating,

$\begin{aligned} A'(x) &= (1 - (g^{-1})'(f(x)) \cdot f'(x)) \cdot f(x) + (x - (g^{-1} \circ f)(x) ) \cdot f'(x) \\ &= \left( 1 - \frac { f'(x) }{g'((g^{-1} \circ f)(x))} \right) \cdot f(x) + (x - (g^{-1} \circ f)(x)) \cdot f'(x).\end{aligned}$

Setting $A'$ yields the desired result.

Problem 2. Suppose further in Problem 1 that $f$ is even. Prove that the $x$ -value $x_0$ that maximises $A$ satisfies the equation

$f(x_0) + x_0 \cdot f'(x_0) = 0$

and the inequality

$2f'(x_0) + x_0 \cdot f''(x_0) < 0.$

Furthermore, the maximum area is given by $A(x_0) = 2x_0 \cdot f(x_0)$ .

(Click for Solution)

Solution. If $f$ is even, then for any $x > 0$ ,

$\displaystyle (g^{-1} \circ f)(x) = g^{-1}(f(x)) = f^{-1}(f(-x)) = -x,$

Furthermore, $f'$ is odd:

$\displaystyle f'(-x)\cdot(-1) = \frac{\mathrm d}{\mathrm dx} (f(-x)) = \frac{\mathrm d}{\mathrm dx}(f(x)) = f'(x),$

so that $f'(-x) = -f'(x)$ . Therefore, for any $x < 0$ , $g(x) = f(x) = f(-x)$ and

$g'(x) = f'(-x) \cdot (-1) = f'(x).$

Hence, for any $x > 0$ ,

$\begin{aligned} (g' \circ g^{-1} \circ f)(x) &= f' ((g^{-1} \circ f)(x)) \\ &= f'(-x) = -f'(x) \neq 0. \end{aligned}$

Substituting into the result in Problem 1,

$\begin{aligned} \left( 1 - \frac { f'(x) }{-f'(x)} \right) \cdot f(x) + (x - (-x)) \cdot f'(x) &= 0, \end{aligned}$

yielding $f(x) + x \cdot f'(x) = 0$ . For the inequality, apply the second derivative test.

Problem 3. Evaluate the maximum area of the rectangle defined in Problem 1 given that $f(x) = 1/( 1+x^2 )$ .

(Click for Solution)

Solution. For $x > 0$ , $f(x) = 1/( 1+x^2 )$ so that

$\displaystyle f'(x) = -\frac 1{( 1+x^2 )^2} \cdot 2x = -\frac{2x}{( 1+x^2 )^2}.$

Since $f$ is even, by Problem 2,

$\displaystyle \begin{aligned} \frac 1{1+x_0^2 } + x_0 \cdot \left( -\frac{2x_0}{ (1+x_0^2)2 } \right) &= 0 \\ 1+x_0^2 + x_0 \cdot (-2x_0) &= 0\\ 1 - x_0^2 &= 0\\ x_0 &= 1, \end{aligned}$