Question-175568

Question Number 175568 by ajfour last updated on 02/Sep/22

Commented by ajfour last updated on 02/Sep/22

$${Find}\:{the}\:{maximum}\:{inner}\:{cone} \\ $$$${volume}\:{if}\:{outer}\:{cone}\:{volume}\:{is} \\ $$$${unity}. \\ $$

Answered by mr W last updated on 02/Sep/22

outer cone: R, H, V inner cone: r, h, v V=((πR^2 H)/3) (=1) v=((πr^2 h)/3) λ=(v/V)=((h/H))((r/R))^2 (r/R)=((H−h)/H)=1−(h/H) with ξ=(h/H) <1 ⇒λ=ξ(1−ξ)^2 (dλ/dξ)=(1−ξ)^2 −2ξ(1−ξ)=0 ⇒ξ=(1/( 3)) λ_(max) =(1/( 3))(1−(1/3))^2 =(4/(27))≈0.148 ✓

$${outer}\:{cone}:\:{R},\:{H},\:{V} \\ $$$${inner}\:{cone}:\:{r},\:{h},\:{v} \\ $$$${V}=\frac{\pi{R}^{\mathrm{2}} {H}}{\mathrm{3}}\:\:\left(=\mathrm{1}\right) \\ $$$${v}=\frac{\pi{r}^{\mathrm{2}} {h}}{\mathrm{3}} \\ $$$$\lambda=\frac{{v}}{{V}}=\left(\frac{{h}}{{H}}\right)\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} \\ $$$$\frac{{r}}{{R}}=\frac{{H}−{h}}{{H}}=\mathrm{1}−\frac{{h}}{{H}} \\ $$$${with}\:\xi=\frac{{h}}{{H}}\:<\mathrm{1} \\ $$$$\Rightarrow\lambda=\xi\left(\mathrm{1}−\xi\right)^{\mathrm{2}} \\ $$$$\frac{{d}\lambda}{{d}\xi}=\left(\mathrm{1}−\xi\right)^{\mathrm{2}} −\mathrm{2}\xi\left(\mathrm{1}−\xi\right)=\mathrm{0}\:\Rightarrow\xi=\frac{\mathrm{1}}{\:\mathrm{3}} \\ $$$$\lambda_{{max}} =\frac{\mathrm{1}}{\:\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{27}}\approx\mathrm{0}.\mathrm{148}\:\checkmark \\ $$

Commented by ajfour last updated on 03/Sep/22