Question

De todos os cones que contém uma esfera de raio R, qual tem o menor volume?

Answer 1

Boa tarde Gregory!


Solução!


$\boxed{V= \dfrac{ \pi .r^{2}.h }{3}}\\\\\ Como ~~o~~ cone~~ pode~~ ser~~ decomposto~~ em ~~dois~~ tri\^agulos~~ retangulos. \\\\\\ Geratriz=L\\\\\ Hipotenusa=(h-r)\\\\\\R=Eesfera\\\\\\\\ Base=r\\\\\ (h-r)^{2}=(l)^{2}+(r)^{2}\\\\\\ h^{2}-2hr+r^{2}=l^{2}+r^{2}\\\\\\ h^{2}-2hr+r^{2}-r^{2} =l^{2}\\\\\\ h^{2}-2hr =l^{2}\\\\\\ \boxed{\sqrt{h^{2}-2hr}=l}$



$Por~~semelhanca!\\\\\\ \dfrac{r}{h}= \dfrac{R}{\sqrt{h^{2}-2hr}} \\\\\\ r= \dfrac{Rh}{\sqrt{h^{2}-2hr}}$


Retomando a formula do volume do cone,vamos substituir o valor do raio.


$V= \dfrac{ \pi .r^{2}.h }{3}}\\\\\\\\ V= \dfrac{ \pi .\bigg(\dfrac{Rh}{\sqrt{h^{2}-2hr}}\bigg)^{2} .h }{3}}\\\\\\\\V= \dfrac{ \pi .\bigg(\dfrac{R^{2}h^{2} }{h^{2}-2hr}\bigg) .h }{3}}\\\\\\\\ V=\bigg(\dfrac{\pi R^{2}h^{3} }{3h^{2} -6hr}\bigg)}\\\\\\\\\ V=\bigg(\dfrac{h(\pi R^{2}h^{2} )}{h(3h-6r)}\bigg) }$




$\boxed{V=\bigg(\dfrac{\pi R^{2}h^{2}}{3h-6r}\bigg) }$

Boa tarde!
Bons estudos!