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Question Number 150688 by ajfour last updated on 14/Aug/21
Commented by ajfour last updated on 14/Aug/21
Find maximum tetrahedron volume.
$${Find}\:{maximum}\:{tetrahedron} \\ $$$${volume}. \\ $$
Answered by mr W last updated on 14/Aug/21
Commented by mr W last updated on 14/Aug/21
((((√6)s)/3)−R)^2 +((s/( (√3))))^2 =R^2 ⇒s=((2(√6)R)/3) V_(tetrahedron) =(s^3 /(6(√2)))=((8(√3)R^3 )/(27)) (V_(tetrahedron) /V_(sphere) )=((8(√3)R^3 )/(27))×(3/(4πR^3 ))=((2(√3))/(9π))≈12%
$$\left(\frac{\sqrt{\mathrm{6}}{s}}{\mathrm{3}}−{R}\right)^{\mathrm{2}} +\left(\frac{{s}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{2}\sqrt{\mathrm{6}}{R}}{\mathrm{3}} \\ $$$${V}_{{tetrahedron}} =\frac{{s}^{\mathrm{3}} }{\mathrm{6}\sqrt{\mathrm{2}}}=\frac{\mathrm{8}\sqrt{\mathrm{3}}{R}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\frac{{V}_{{tetrahedron}} }{{V}_{{sphere}} }=\frac{\mathrm{8}\sqrt{\mathrm{3}}{R}^{\mathrm{3}} }{\mathrm{27}}×\frac{\mathrm{3}}{\mathrm{4}\pi{R}^{\mathrm{3}} }=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}\pi}\approx\mathrm{12\%} \\ $$
Commented by ajfour last updated on 14/Aug/21
sir, i think, i couldn′t convey my question clearly. i meant a tetrahedron excavated out of a sphere of radius R, with one vertex at centre of sphere, while rest three on the surface of sphere. It should be of max volume and not necessarily regular. V=(1/3)(((√3)/4)s^2 )(√(R^2 −((s/(2cos 30°)))^2 )) V^( 2) =(((s^2 )^2 )/(48))(R^2 −(s^2 /3)) (dV^( 2) /ds^2 )=0 ⇒ 2s^2 (R^2 −(s^2 /3))−(((s^2 )^2 )/3)=0 ⇒ s^2 =2R^2 V_(max) =(R^2 /( 2(√3)))((R/( (√3))))=(R^3 /6)
$${sir},\:{i}\:{think},\:{i}\:{couldn}'{t}\:{convey} \\ $$$${my}\:{question}\:{clearly}. \\ $$$${i}\:{meant}\:{a}\:{tetrahedron} \\ $$$${excavated}\:{out}\:{of}\:{a}\:{sphere}\:{of} \\ $$$${radius}\:{R},\:{with}\:{one}\:{vertex} \\ $$$${at}\:{centre}\:{of}\:{sphere},\:{while} \\ $$$${rest}\:{three}\:{on}\:{the}\:{surface}\:{of} \\ $$$${sphere}.\:{It}\:{should}\:{be}\:{of}\:{max} \\ $$$${volume}\:{and}\:{not}\:{necessarily} \\ $$$${regular}.\: \\ $$$$\:{V}=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{s}^{\mathrm{2}} \right)\sqrt{{R}^{\mathrm{2}} −\left(\frac{{s}}{\mathrm{2cos}\:\mathrm{30}°}\right)^{\mathrm{2}} } \\ $$$${V}^{\:\mathrm{2}} =\frac{\left({s}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{48}}\left({R}^{\mathrm{2}} −\frac{{s}^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$\frac{{dV}^{\:\mathrm{2}} }{{ds}^{\mathrm{2}} }=\mathrm{0}\:\:\Rightarrow\:\mathrm{2}{s}^{\mathrm{2}} \left({R}^{\mathrm{2}} −\frac{{s}^{\mathrm{2}} }{\mathrm{3}}\right)−\frac{\left({s}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{s}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} \\ $$$${V}_{{max}} =\frac{{R}^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{{R}}{\:\sqrt{\mathrm{3}}}\right)=\frac{{R}^{\mathrm{3}} }{\mathrm{6}} \\ $$
Commented by mr W last updated on 14/Aug/21
i see now.
$${i}\:{see}\:{now}. \\ $$
Answered by ajfour last updated on 14/Aug/21
Rsin θcos ((π/6))=(s/2) s=R(√3)sin θ V=(1/3)(((√3)/4)s^2 )Rcos θ =(1/( 3))(((√3)/4))(3R^2 sin^2 θ)Rcos θ (dV/dθ)=((√3)/4)R^3 (2sin θcos^2 θ−sin^3 θ) (dV/dθ)=0 ⇒ tan θ=0, (√2) V_(max) =(((√3)R^3 )/4)((2/3))((1/( (√3)))) V_(max) =(R^3 /6) (V_(max) /V_(sphere) )=(1/6)((3/(4π)))=(1/(8π)) (but this is very less! something wrong, i guess!)
$${R}\mathrm{sin}\:\theta\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}\right)=\frac{{s}}{\mathrm{2}} \\ $$$${s}={R}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\:\: \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{s}^{\mathrm{2}} \right){R}\mathrm{cos}\:\theta \\ $$$$\:\:=\frac{\mathrm{1}}{\:\mathrm{3}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)\left(\mathrm{3}{R}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right){R}\mathrm{cos}\:\theta \\ $$$$\frac{{dV}}{{d}\theta}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{R}^{\mathrm{3}} \left(\mathrm{2sin}\:\theta\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{3}} \theta\right) \\ $$$$\frac{{dV}}{{d}\theta}=\mathrm{0}\:\:\:\Rightarrow\:\:\mathrm{tan}\:\theta=\mathrm{0},\:\sqrt{\mathrm{2}} \\ $$$${V}_{{max}} =\frac{\sqrt{\mathrm{3}}{R}^{\mathrm{3}} }{\mathrm{4}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${V}_{{max}} =\frac{{R}^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\frac{{V}_{{max}} }{{V}_{{sphere}} }=\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{3}}{\mathrm{4}\pi}\right)=\frac{\mathrm{1}}{\mathrm{8}\pi} \\ $$$$\left({but}\:{this}\:{is}\:{very}\:{less}!\right. \\ $$$$\left.{something}\:{wrong},\:{i}\:{guess}!\right) \\ $$
Commented by mr W last updated on 14/Aug/21
it′s correct sir!
$${it}'{s}\:{correct}\:{sir}! \\ $$
Answered by mr W last updated on 14/Aug/21
Commented by mr W last updated on 14/Aug/21
r=(√(R^2 −h^2 )) a=side length of base triangle a=(√3)r=(√(3(R^2 −h^2 ))) A=(((√3)a^2 )/4)=((3(√3)(R^2 −h^2 ))/4) V=((Ah)/3)=(((√3)(R^2 −h^2 )h)/4) (dV/dh)=0 ⇒R^2 −3h^2 =0 ⇒h=(R/( (√3))) V_(max) =(((√3)(R^2 −(R^2 /3))×(R/( (√3))))/4)=(R^3 /6)
$${r}=\sqrt{{R}^{\mathrm{2}} −{h}^{\mathrm{2}} } \\ $$$${a}={side}\:{length}\:{of}\:{base}\:{triangle} \\ $$$${a}=\sqrt{\mathrm{3}}{r}=\sqrt{\mathrm{3}\left({R}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)} \\ $$$${A}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}\left({R}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)}{\mathrm{4}} \\ $$$${V}=\frac{{Ah}}{\mathrm{3}}=\frac{\sqrt{\mathrm{3}}\left({R}^{\mathrm{2}} −{h}^{\mathrm{2}} \right){h}}{\mathrm{4}} \\ $$$$\frac{{dV}}{{dh}}=\mathrm{0}\:\Rightarrow{R}^{\mathrm{2}} −\mathrm{3}{h}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{h}=\frac{{R}}{\:\sqrt{\mathrm{3}}} \\ $$$${V}_{{max}} =\frac{\sqrt{\mathrm{3}}\left({R}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{3}}\right)×\frac{{R}}{\:\sqrt{\mathrm{3}}}}{\mathrm{4}}=\frac{{R}^{\mathrm{3}} }{\mathrm{6}} \\ $$
Commented by ajfour last updated on 14/Aug/21
thanks sir.
$${thanks}\:{sir}. \\ $$

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