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Mathematics Question on Applications of Derivatives

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 827\frac{8}{27} of the volume of the sphere.

Answer

Let rr and h be the radius and height of the cone respectively inscribed in a sphere of

radius R.

Let VV be the volume of the cone.

Then,V=13πr2h,V=\frac{1}{3}\pi r^{2}h

Height of the cone is given by,

h=R+AB=R+R2−r2[ABC  is  a  right  triangle]h=R+AB=R+\sqrt{R^{2}-r^{2}} [ABC \space is\space a\space right\space triangle]

∴V=13πr2h(R+R2−r2)∴V=\frac{1}{3}\pi r^{2}h(R+\sqrt{R^{2}-r^{2})}

=13πr2hR+13πr2R2−r2=\frac{1}{3}\pi r^{2}hR+\frac{1}{3}\pi r^{2}\sqrt{R^{2}-r^{2}}

∴\frac{dV}{dr}=$$\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{R^{2}-r^{2}}+\frac{1}{3}\pi r^{2}.\frac{(-2r)}{2\sqrt{R^{2}-r^{2}}}

=23πrR+23πrR2−r2−13πr3R2−r2=\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{R^{2}-r^{2}}-\frac{1}{3}\pi \frac{r^{3}}{\sqrt{R^{2}-r^{2}}}

=23πrR+2πr(R2−r2)−πr33R2−r2=\frac{2}{3}\pi rR+\frac{2\pi r(R^{2}-r^{2})-\pi r^{3}}{3\sqrt{R^{2}-r^{2}}}

=23πrR+2πrR2−3πr33R2−r2=\frac{2}{3}\pi rR+\frac{2\pi rR^{2}-3\pi r^{3}}{3\sqrt{R^{2}-r^{2}}}

d2Vdr2\frac{d^{2}V}{dr^{2}}=2πR3\frac{2\pi R}{3}+3R2−r2(2πR2−9πr2)−(2πrR2−3πr3).(−2r)6R2−r29(R2−r2)\frac{3\sqrt{R^{2}-r^{2}}(2\pi R^{2}-9\pi r^{2})-(2\pi rR^{2}-3\pi r^{3}).\frac{(-2r)}{{6\sqrt{R^{2}-r^{2}}}}}{9(R^{2}-r^{2})}

=\frac{2}{3}\pi R$$+\frac{9(R^{2}-r^{2})(2\pi R^{2}-9\pi r^{2})+2\pi r^{2}+3\pi r^{4}}{27(R^{2}-r^{2})\frac{3}{2}}

Now,\frac{dV}{dr}$$=0⇒$$\pi 23rR=3πr3−2πrR23R2−r2\frac{3\pi r^{3}-2\pi rR^{2}}{3\sqrt{R^{2}-r^{2}}}

⇒2R={3r^{2}-2R2}{\sqrt{R^{2}-r^{2}}}$$⇒2R\sqrt{R^{2}-r^{2}}=3r^{2}-2R^{2}

⇒4R2(R2−r2)=(3r2−2R2)2⇒4R^{2}(R^{2}-r^{2})=(3r^{2}-2R^{2})^{2}

⇒4R4−4R2r2=9r4+4R4−12r2R2⇒4R^{4}-4R^{2}r^{2}=9r^{4}+4R^{4}-12r^{2}R^{2}

⇒9r4=8R2r2⇒9r^{4}=8R^{2}r^{2}

⇒r2=89R2⇒r^{2}=\frac{8}{9R^{2}}

When r2=89R2,r^{2}=\frac{8}{9R^{2}},thend2Vdr2<0.\frac{d^{2}V}{dr^{2}}<0.

⧠By second derivative test,the volume of the cone is the maximum when r2=89R2.r^{2}=\frac{8}{9R^{2}}.

When r2=89R2r^{2}=\frac{8}{9R^{2}},h=R+\sqrt{R^{2}-\frac{8}{9}R^{2}}$$=R+\sqrt{\frac{1}{9R^{2}}}=R+\frac{R}{3}=\frac{4}{3R.}

Therefore,

=13π(89R2)(43R)=\frac{1}{3}\pi (\frac{8}{9}R^{2})(\frac{4}{3}R)

=827(43πR3)=\frac{8}{27}(\frac{4}{3}\pi R^{3})

=827×=\frac{8}{27}\times(Volume of sphere)

Hence,the volume of the largest cone that can be inscribed in the sphere is827\frac{8}{27}

the volume of the sphere.