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Question Number 138388 by ajfour last updated on 13/Apr/21
Find maximum volume of a  cylinder within a unit cube   whose axis is along a diagonal  of the cube.
$${Find}\:{maximum}\:{volume}\:{of}\:{a} \\ $$$${cylinder}\:{within}\:{a}\:{unit}\:{cube}\: \\ $$$${whose}\:{axis}\:{is}\:{along}\:{a}\:{diagonal} \\ $$$${of}\:{the}\:{cube}. \\ $$
Answered by mr W last updated on 13/Apr/21
r=radius of cylinder  l=length of cylinder  a=edge length of cube=1  l=(√3)a−2(√2)r  V=πr^2 l=πr^2 ((√3)a−2(√2)r)  (dV/dr)=π(2(√3)ar−6(√2)r^2 )=0  ⇒r_m =(a/( (√6)))  V_(max) =π×(a^2 /6)((√3)a−((2(√2)a)/( (√6))))=(((√3)πa^3 )/(18))
$${r}={radius}\:{of}\:{cylinder} \\ $$$${l}={length}\:{of}\:{cylinder} \\ $$$${a}={edge}\:{length}\:{of}\:{cube}=\mathrm{1} \\ $$$${l}=\sqrt{\mathrm{3}}{a}−\mathrm{2}\sqrt{\mathrm{2}}{r} \\ $$$${V}=\pi{r}^{\mathrm{2}} {l}=\pi{r}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}{a}−\mathrm{2}\sqrt{\mathrm{2}}{r}\right) \\ $$$$\frac{{dV}}{{dr}}=\pi\left(\mathrm{2}\sqrt{\mathrm{3}}{ar}−\mathrm{6}\sqrt{\mathrm{2}}{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{r}_{{m}} =\frac{{a}}{\:\sqrt{\mathrm{6}}} \\ $$$${V}_{{max}} =\pi×\frac{{a}^{\mathrm{2}} }{\mathrm{6}}\left(\sqrt{\mathrm{3}}{a}−\frac{\mathrm{2}\sqrt{\mathrm{2}}{a}}{\:\sqrt{\mathrm{6}}}\right)=\frac{\sqrt{\mathrm{3}}\pi{a}^{\mathrm{3}} }{\mathrm{18}} \\ $$
Commented by mr W last updated on 13/Apr/21
fine! it′s more elegant using vectors.
$${fine}!\:{it}'{s}\:{more}\:{elegant}\:{using}\:{vectors}. \\ $$
Commented by ajfour last updated on 13/Apr/21
yes sir, thank you, understood.  cos φ=(((i+j+k))/( (√3)))∙(((i+j))/( (√2)))=((√2)/( (√3)))  tan φ=(r/b) =(1/( (√2)))  ⇒  b=r(√2)  2b+l=a(√3)  2r(√2)+l=a(√3)  l=a(√3)−2r(√2)  ....  ....  r_m =(a/( (√6)))   ,  V_(max) =(((√3)πa^3 )/(18))
$${yes}\:{sir},\:{thank}\:{you},\:{understood}. \\ $$$$\mathrm{cos}\:\phi=\frac{\left({i}+{j}+{k}\right)}{\:\sqrt{\mathrm{3}}}\centerdot\frac{\left({i}+{j}\right)}{\:\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{tan}\:\phi=\frac{{r}}{{b}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\Rightarrow\:\:{b}={r}\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}{b}+{l}={a}\sqrt{\mathrm{3}} \\ $$$$\mathrm{2}{r}\sqrt{\mathrm{2}}+{l}={a}\sqrt{\mathrm{3}} \\ $$$${l}={a}\sqrt{\mathrm{3}}−\mathrm{2}{r}\sqrt{\mathrm{2}} \\ $$$$…. \\ $$$$…. \\ $$$${r}_{{m}} =\frac{{a}}{\:\sqrt{\mathrm{6}}}\:\:\:,\:\:{V}_{{max}} =\frac{\sqrt{\mathrm{3}}\pi{a}^{\mathrm{3}} }{\mathrm{18}} \\ $$
Commented by mr W last updated on 14/Apr/21
can you prove Q138519 using vector  method?  i found that by accident, but  have no proof yet.
$${can}\:{you}\:{prove}\:{Q}\mathrm{138519}\:{using}\:{vector} \\ $$$${method}?\:\:{i}\:{found}\:{that}\:{by}\:{accident},\:{but} \\ $$$${have}\:{no}\:{proof}\:{yet}. \\ $$
Answered by mr W last updated on 14/Apr/21

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