Question Number 73828 by ajfour last updated on 16/Nov/19
Commented by ajfour last updated on 17/Nov/19
$${If}\:{on}\:{each}\:{face}\:{of}\:{the}\:{larger}\:{cube} \\ $$$${there}\:{is}\:\left({at}\:{least}\right)\:{one}\:{corner}\:{of}\: \\ $$$${the}\:{inner}\:{cube},\:{then} \\ $$$${find}\:{minimum}\:{value}\:{of}\:{ratio}\:{r}. \\ $$$$\:{r}=\frac{{s}}{{a}}\:=\:\frac{{edge}\:{length}\:{of}\:{inner}\:{cube}}{{edge}\:{length}\:{of}\:{outer}\:{cube}}\:. \\ $$
Commented by ajfour last updated on 17/Nov/19
$${MjS}\:{Sir}\:\&\:'{Powerful}\:{Mind}'\:{Sir} \\ $$$${please}\:{attempt}\:{this}\:{question}.. \\ $$
Commented by mind is power last updated on 17/Nov/19
$$\mathrm{i}\:\mathrm{will}\:\mathrm{translste}\:\mathrm{in}\:\mathrm{french}\:\mathrm{and}\:\mathrm{i}\:\mathrm{will}\:\mathrm{try}\:,\mathrm{my}\:\mathrm{english}\:\mathrm{is}\:\mathrm{limited}\:! \\ $$
Answered by ajfour last updated on 17/Nov/19
Commented by ajfour last updated on 17/Nov/19
![Let A and B do not touch outer cube walls.Outer cube side=1 C(0,y,z) ; G(a,b,0); E(h,0,k) A(((h+a)/2), (b/2)−y , (k/2)−z) h^2 +y^2 +(k−z)^2 =s^2 ( =CE^2 ) ..(i) a^2 +(b−y)^2 +z^2 =s^2 (=CG^2 ) ..(ii) CE^(→) ⊥ CG^(→) ⇒ ah+(y−b)y+(z−k)z=0 ...(iii) CB^(→) = CE^(→) ×CG^(→) = determinant ((i,j,k),(h,(−y),(k−z)),(a,(b−y),(−z))) ={yz−(b−y)(k−z)]i^ +{a(k−z)+hz}j^ +{h(b−y)+ay}k^ x_D =x_A +(CB^(→) )_x = 1 ⇒ ((h+a)/2)+yz−(b−y)(k−z)=1 (iv) y_F = y_G +(CB^(→) )_y = 1 ⇒ b+a(k−z)+hz=1 (v) z_H = z_E +(CB^(→) )_z = 1 ⇒ k+h(b−y)+ay=1 (vi) unknowns y,z,a,b,h,k,s (7) equations (i)→(vi) ⇒ s=f(y) & then (ds/dy)=0 gives hopefully s_(min) .](Q74012.png)
$${Let}\:{A}\:{and}\:{B}\:{do}\:{not}\:{touch}\:{outer} \\ $$$${cube}\:{walls}.{Outer}\:{cube}\:{side}=\mathrm{1} \\ $$$${C}\left(\mathrm{0},{y},{z}\right)\:\:;\:\:{G}\left({a},{b},\mathrm{0}\right);\:\:{E}\left({h},\mathrm{0},{k}\right) \\ $$$${A}\left(\frac{{h}+{a}}{\mathrm{2}},\:\frac{{b}}{\mathrm{2}}−{y}\:,\:\frac{{k}}{\mathrm{2}}−{z}\right) \\ $$$${h}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({k}−{z}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \:\:\left(\:={CE}^{\mathrm{2}} \right)\:\:..\left({i}\right) \\ $$$${a}^{\mathrm{2}} +\left({b}−{y}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} ={s}^{\mathrm{2}} \:\:\:\left(={CG}^{\mathrm{2}} \right)\:\:..\left({ii}\right) \\ $$$$\overset{\rightarrow} {{CE}}\:\bot\:\overset{\rightarrow} {{CG}}\:\:\:\Rightarrow \\ $$$${ah}+\left({y}−{b}\right){y}+\left({z}−{k}\right){z}=\mathrm{0}\:\:\:\:\:\:\:...\left({iii}\right) \\ $$$$\overset{\rightarrow} {{CB}}\:=\:\overset{\rightarrow} {{CE}}×\overset{\rightarrow} {{CG}} \\ $$$$\:\:\:\:\:\:\:\:=\begin{vmatrix}{{i}}&{{j}}&{{k}}\\{{h}}&{−{y}}&{{k}−{z}}\\{{a}}&{{b}−{y}}&{−{z}}\end{vmatrix} \\ $$$$\:\:\:\:\:=\left\{{yz}−\left({b}−{y}\right)\left({k}−{z}\right)\right]\hat {{i}} \\ $$$$\:+\left\{{a}\left({k}−{z}\right)+{hz}\right\}\hat {{j}}+\left\{{h}\left({b}−{y}\right)+{ay}\right\}\hat {{k}} \\ $$$$ \\ $$$${x}_{{D}} ={x}_{{A}} +\left(\overset{\rightarrow} {{CB}}\right)_{{x}} =\:\mathrm{1}\:\:\:\:\Rightarrow \\ $$$$\frac{{h}+{a}}{\mathrm{2}}+{yz}−\left({b}−{y}\right)\left({k}−{z}\right)=\mathrm{1}\:\:\:\:\:\left({iv}\right) \\ $$$${y}_{{F}} =\:{y}_{{G}} +\left(\overset{\rightarrow} {{CB}}\right)_{{y}} =\:\mathrm{1}\:\:\:\:\Rightarrow \\ $$$$\:\:{b}+{a}\left({k}−{z}\right)+{hz}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({v}\right) \\ $$$$ \\ $$$$\:{z}_{{H}} =\:{z}_{{E}} +\left(\overset{\rightarrow} {{CB}}\right)_{{z}} \:=\:\mathrm{1}\:\:\:\Rightarrow \\ $$$$\:{k}+{h}\left({b}−{y}\right)+{ay}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({vi}\right) \\ $$$$ \\ $$$${unknowns}\:{y},{z},{a},{b},{h},{k},{s}\:\:\:\left(\mathrm{7}\right) \\ $$$${equations}\:\:\left({i}\right)\rightarrow\left({vi}\right) \\ $$$$\Rightarrow\:\:{s}={f}\left({y}\right)\:\:\&\:\:{then}\:\:\frac{{ds}}{{dy}}=\mathrm{0}\:\:{gives} \\ $$$${hopefully}\:\boldsymbol{{s}}_{{min}} . \\ $$
Commented by ajfour last updated on 18/Nov/19
Commented by ajfour last updated on 18/Nov/19
