Maxima & Minima

Show that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Comments

answer


AD=H CD=R FG=r FH=g

$$Area(r,g)=2\pi r g$$
$$\frac{r}{R}=\frac{H-g}{H}$$

so

$$r=\frac{R(H-g)}{H}$$
$$area_{cyl}(g)=2\pi\frac{R}{H}g(H-g)$$

This is a function of second degree in g, which have maximal value for

$$2g=H$$

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