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Question-49696

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Question Number 49696 by ajfour last updated on 09/Dec/18
Commented by ajfour last updated on 09/Dec/18
Find parameters of ellipse within the box and touching all its six faces at A, B, C, D, E, and F and (may be for uniqueness) of maximum area.
$${Find}\:{parameters}\:{of}\:{ellipse}\:{within} \\ $$$${the}\:{box}\:{and}\:{touching}\:{all}\:{its}\:{six} \\ $$$${faces}\:{at}\:{A},\:{B},\:{C},\:{D},\:{E},\:{and}\:{F}\:\:{and} \\ $$$$\left({may}\:{be}\:{for}\:{uniqueness}\right)\:{of} \\ $$$${maximum}\:{area}. \\ $$
Answered by mr W last updated on 09/Dec/18
Commented by mr W last updated on 09/Dec/18
projection of the ellipse in xy−plane is also an ellipse, let′s say with parameters p and b: (x^2 /p^2 )+(y^2 /b^2 )=1 eqn. of tangent line: x+y−(l/( (√2)))=0 ⇒p^2 +b^2 =((l/( (√2))))^2 =(l^2 /2) ⇒b^2 =(l^2 /2)−p^2 (2a)^2 =(2p)^2 +h^2 ⇒a^2 =p^2 +(h^2 /4) let T=a^2 b^2 =(p^2 +(h^2 /4))((l^2 /2)−p^2 ) such that the ellipse has maximum area, (dT/dp)=0 (dT/dp)=(p^2 +(h^2 /4))(−2p)+2p((l^2 /2)−p^2 )=0 ⇒p^2 =(1/8)(2l^2 −h^2 ) ⇒a^2 =(1/8)(2l^2 −h^2 )+(h^2 /4)=(1/8)(2l^2 +h^2 ) ⇒a=(1/2)(√(l^2 +(h^2 /2))) ⇒b^2 =(l^2 /2)−(1/8)(2l^2 −h^2 )=(1/8)(2l^2 +h^2 ) ⇒b=(1/2)(√(l^2 +(h^2 /2)))=a i.e. the ellipse with maximum area is a circle with R=a=b=(1/2)(√(l^2 +(h^2 /2))) max.A=πR^2 =((π(2l^2 +h^2 ))/8)
$${projection}\:{of}\:{the}\:{ellipse}\:{in}\:{xy}−{plane} \\ $$$${is}\:{also}\:{an}\:{ellipse},\:{let}'{s}\:{say}\:{with} \\ $$$${parameters}\:{p}\:{and}\:{b}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{p}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${eqn}.\:{of}\:{tangent}\:{line}: \\ $$$${x}+{y}−\frac{{l}}{\:\sqrt{\mathrm{2}}}=\mathrm{0} \\ $$$$\Rightarrow{p}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\frac{{l}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{{l}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\frac{{l}^{\mathrm{2}} }{\mathrm{2}}−{p}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\left(\mathrm{2}{p}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={p}^{\mathrm{2}} +\frac{{h}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${let}\:{T}={a}^{\mathrm{2}} {b}^{\mathrm{2}} =\left({p}^{\mathrm{2}} +\frac{{h}^{\mathrm{2}} }{\mathrm{4}}\right)\left(\frac{{l}^{\mathrm{2}} }{\mathrm{2}}−{p}^{\mathrm{2}} \right) \\ $$$${such}\:{that}\:{the}\:{ellipse}\:{has}\:{maximum} \\ $$$${area},\:\frac{{dT}}{{dp}}=\mathrm{0} \\ $$$$\frac{{dT}}{{dp}}=\left({p}^{\mathrm{2}} +\frac{{h}^{\mathrm{2}} }{\mathrm{4}}\right)\left(−\mathrm{2}{p}\right)+\mathrm{2}{p}\left(\frac{{l}^{\mathrm{2}} }{\mathrm{2}}−{p}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{l}^{\mathrm{2}} +{h}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{l}^{\mathrm{2}} +\frac{{h}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\frac{{l}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{l}^{\mathrm{2}} +{h}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{l}^{\mathrm{2}} +\frac{{h}^{\mathrm{2}} }{\mathrm{2}}}={a} \\ $$$${i}.{e}.\:{the}\:{ellipse}\:{with}\:{maximum}\:{area}\:{is} \\ $$$${a}\:{circle}\:{with}\:{R}={a}={b}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{l}^{\mathrm{2}} +\frac{{h}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${max}.{A}=\pi{R}^{\mathrm{2}} =\frac{\pi\left(\mathrm{2}{l}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)}{\mathrm{8}} \\ $$
Commented by ajfour last updated on 09/Dec/18
GREAT SiR! Thanks a lot.
$$\mathbb{GREAT}\:\mathcal{S}{i}\mathcal{R}!\:\:{Thanks}\:{a}\:{lot}. \\ $$

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