Page:Calculus Made Easy.pdf/136

This page has been proofread, but needs to be validated.

116

Calculus Made Easy

(3) Find the maxima and minima of $y={\dfrac {x-1}{x^{2}+2}}$ .

${\frac {dy}{dx}}={\frac {(x^{2}+2)\times 1-(x-1)\times 2x}{(x^{2}+2)^{2}}}={\frac {2x-x^{2}+2}{(x^{2}+2)^{2}}}=0$ ;

or $x^{2}-2x-2=0$ , whose solutions are $x=+2.73$ and $x=-0.73$ .

${\begin{aligned}{\dfrac {d^{2}y}{dx^{2}}}&=-{\frac {(x^{2}+2)^{2}\times (2x-2)-(x^{2}-2x-2)(4x^{3}+8x)}{(x^{2}+2)^{4}}}\\&=-{\frac {2x^{5}-6x^{4}-8x^{3}-8x^{2}-24x+8}{(x^{2}+2)^{4}}}\end{aligned}}$ .

The denominator is always positive, so it is sufficient to ascertain the sign of the numerator.

If we put $x=2.73$ , the numerator is negative; the maximum, $y=0.183$ .

If we put $x=-0.73$ , the numerator is positive; the minimum, $y=-0.683$ .

(4) The expense $C$ of handling the products of a certain factory varies with the weekly output $P$ according to the relation $C=aP+{\dfrac {b}{c+P}}+d$ , where $a$ , $b$ , $c$ , $d$ are positive constants. For what output will the expense be least?

${\dfrac {dC}{dP}}=a-{\frac {b}{(c+P)^{2}}}=0\quad$ for maximum or minimum;

hence $a={\dfrac {b}{(c+P)^{2}}}$ and $P=\pm {\sqrt {\dfrac {b}{a}}}-c$ .

As the output cannot be negative, $P=+{\sqrt {\dfrac {b}{a}}}-c$ .

This article is issued from Wikisource. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.