Page:Calculus Made Easy.pdf/125

 You will get

For maximum or minimum we must have

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, $$H=\sqrt{l^2-R^2}$$.

Proceeding as in the previous problem, we get

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.