Page:Calculus Made Easy.pdf/125

You will get

${\dfrac {dS}{dx}}=x\times -{\dfrac {x}{\sqrt {4R^{2}-x^{2}}}}+{\sqrt {4R^{2}-x^{2}}}={\dfrac {4R^{2}-2x^{2}}{\sqrt {4R^{2}-x^{2}}}}$.

For maximum or minimum we must have

${\dfrac {4R^{2}-2x^{2}}{\sqrt {4R^{2}-x^{2}}}}=0$;

that is, $4R^{2}-2x^{2}=0$ and $x=R{\sqrt {2}}$.

The other side $={\sqrt {4R^{2}-2R^{2}}}=R{\sqrt {2}}$; the two sides are equal; the figure is a square the side of which is equal to the diagonal of the square constructed on the radius. In this case it is, of course, a maximum with which we are dealing.

(2) What is the radius of the opening of a conical vessel the sloping side of which has a length $l$ when the capacity of the vessel is greatest?

If $R$ be the radius and $H$ the corresponding height, $H={\sqrt {l^{2}-R^{2}}}$.

Volume $V=\pi R^{2}\times {\dfrac {H}{3}}=\pi R^{2}\times {\dfrac {\sqrt {l^{2}-R^{2}}}{3}}$.

Proceeding as in the previous problem, we get

${\dfrac {dV}{dR}}=\pi R^{2}\times -{\dfrac {R}{3{\sqrt {l^{2}-R^{2}}}}}+{\dfrac {2\pi R}{3}}{\sqrt {l^{2}-R^{2}}}$

$={\dfrac {2\pi R(l^{2}-R^{2})-\pi R^{3}}{3{\sqrt {l^{2}-R^{2}}}}}=0$

for maximum or minimum.

Or, $2\pi R(l^{2}-R^{2})-\pi R^{2}=0$, and $R=l{\sqrt {\tfrac {2}{3}}}$ for a maximum, obviously.