Page:Calculus Made Easy.pdf/124

If the curve is such that there is no place that is a maximum or minimum, the process of equating to zero will yield an impossible result. For instance:

Let

$y=ax^{3}+bx+c$.

Then

${\frac {dy}{dx}}=3ax^{2}+b$.

Equating this to zero, we get $3ax^{2}+b=0$.

$x^{2}={\frac {-b}{3a}}$, and $x={\sqrt {\frac {-b}{3a}}}$, which is impossible.

Therefore $y$ has no maximum nor minimum.

A few more worked examples will enable you to thoroughly master this most interesting and useful application of the calculus.

(1) What are the sides of the rectangle of maximum area inscribed in a circle of radius $R$?

If one side be called $x$,

the other side$={\sqrt {({\text{diagonal}})^{2}-x^{2}}}$;

and as the diagonal of the rectangle is necessarily a diameter, the other side $={\sqrt {4R^{2}-x^{2}}}$.

Then, area of rectangle $S=x{\sqrt {4R^{2}-x^{2}}}$,

${\frac {dS}{dx}}=x\times {\dfrac {d\left({\sqrt {4R^{2}-x^{2}}}\,\right)}{dx}}+{\sqrt {4R^{2}-x^{2}}}\times {\dfrac {d(x)}{dx}}$.

If you have forgotten how to differentiate ${\sqrt {4R^{2}-x^{2}}}$, here is a hint: write $4R^{2}-x^{2}=w$ and $y={\sqrt {w}}$ and seek ${\dfrac {dy}{dw}}$ and ${\dfrac {dw}{dx}}$; fight it out, and only if you can’t get on refer to page 67.