Question Number 165600 by mr W last updated on 05/Feb/22
Commented by mr W last updated on 05/Feb/22
$${attempt}\:{to}\:{Q}\mathrm{165565} \\ $$
Commented by mr W last updated on 05/Feb/22
Commented by mr W last updated on 06/Feb/22
![let μ=(b/a)≤1 A((a/(tan θ)), a)=(ma, a) B(b, y_B ) (y_B −a)^2 +((a/(tan θ))−b)^2 =(a+b)^2 (y_B −a)^2 =a^2 (1+(1/(tan θ)))(2μ+1−(1/(tan θ))) let m=(1/(tan θ)) y_B =a+a(√((1+m)(2μ+1−m))) tan θ≥(a/(a+2b))=(1/(2μ+1)) θ≤(π/4) ⇒1≤m≤2μ+1 x_P =(a/(tan θ))+a sin α=a(m+sin α) y_P =a−a cos α=a(1−cos α) x_Q =b−b cos β=aμ(1−cos β) y_Q =y_B +b sin β=a[1+(√((1+m)(2μ+1−m)))+μ sin β] area of triangle OPQ: Δ=((x_Q y_Q )/2)+(((x_P −x_Q )(y_P +y_Q ))/2)−((x_P y_P )/2) 2Δ=x_P y_Q −x_Q y_P 2Δ=a^2 (m+sin α)(1+(√((1+m)(2μ+1−m)))+μ sin β)−a^2 μ(1−cos β)(1−cos α) ((2Δ)/(μa^2 ))=(m+sin α)(((1+(√((1+m)(2μ+1−m))))/μ)+sin β)−(1−cos β)(1−cos α) let Φ=((2Δ)/(μa^2 )), λ=((1+(√((1+m)(2μ+1−m))))/μ) Φ=(m+sin α)(λ+sin β)−(1−cos β)(1−cos α) Φ=mλ−1+λ sin α+cos α+m sin β+cos β−cos (α+β) Φ=mλ−1+(√(1+λ^2 )) cos (α−tan^(−1) λ)+(√(1+m^2 )) cos (β−tan^(−1) m)−cos (α+β) Φ is a function of three variables: m, α, β. i′ll try in geometric way: for maiximum area of triangle OPQ, tangent at Q must be // OP: (1/(tan β))=(y_P /x_P )=((1−cos α)/(m+sin α)) ⇒((cos β)/(sin β))=((1−cos α)/(m+sin α)) m cos β−sin β+sin α cos β+cos α sin β=0 m cos β−sin β+sin (α+β)=0 ⇒(√(1+m^2 )) sin (β−tan^(−1) m)=sin (α+β) (we can get the same also from (∂Φ/∂β)=0) tangent at P must be // OQ: tan α=(y_Q /x_Q )=((1+(√((1+m)(2μ+1−m)))+μ sin β)/(μ(1−cos β))) ⇒((sin α)/(cos α))=((λ+sin β)/(1−cos β)) λ cos α−sin α+cos α sin β+sin α cos β=0 λ cos α−sin α+sin (α+β)=0 ⇒(√(1+λ^2 )) sin (α−tan^(−1) λ)=sin (α+β) (we can get the same also from (∂Φ/∂α)=0) we can see that (π/2)<α+β<π in case of that Δ is maximum. say (√(1+λ^2 )) sin (α−tan^(−1) λ)=(√(1+m^2 )) sin (β−tan^(−1) m)=sin (α+β)=ξ ⇒α=tan^(−1) λ+sin^(−1) (ξ/( (√(1+λ^2 )))) ⇒β=tan^(−1) m+sin^(−1) (ξ/( (√(1+m^2 )))) ⇒α+β=π−sin^(−1) ξ determinant (((sin^(−1) ξ+sin^(−1) (ξ/( (√(1+λ^2 ))))+sin^(−1) (ξ/( (√(1+m^2 ))))=π−tan^(−1) λ−tan^(−1) m))) ...(I) (√(1+λ^2 )) sin (α−tan^(−1) λ)=ξ ⇒cos (α−tan^(−1) λ)=(√(1−(ξ^2 /(1+λ^2 )))) (√(1+m^2 )) sin (β−tan^(−1) m)=ξ ⇒cos (β−tan^(−1) m)=(√(1−(ξ^2 /(1+m^2 )))) sin (α+β)=ξ ⇒cos (α+β)=−(√(1−ξ^2 )) Φ=mλ−1+(√(1+λ^2 )) cos (α−tan^(−1) λ)+(√(1+m^2 )) cos (β−tan^(−1) m)−cos (α+β) determinant (((Φ=mλ−1+(√(1+λ^2 −ξ^2 ))+(√(1+m^2 −ξ^2 )) +(√(1−ξ^2 ))))) ...(II)](https://www.tinkutara.com/question/Q165603.png)
$${let}\:\mu=\frac{{b}}{{a}}\leqslant\mathrm{1} \\ $$$${A}\left(\frac{{a}}{\mathrm{tan}\:\theta},\:{a}\right)=\left({ma},\:{a}\right) \\ $$$${B}\left({b},\:{y}_{{B}} \right) \\ $$$$\left({y}_{{B}} −{a}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{tan}\:\theta}−{b}\right)^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\left({y}_{{B}} −{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\right)\left(\mathrm{2}\mu+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\right) \\ $$$${let}\:{m}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta} \\ $$$${y}_{{B}} ={a}+{a}\sqrt{\left(\mathrm{1}+{m}\right)\left(\mathrm{2}\mu+\mathrm{1}−{m}\right)} \\ $$$$\mathrm{tan}\:\theta\geqslant\frac{{a}}{{a}+\mathrm{2}{b}}=\frac{\mathrm{1}}{\mathrm{2}\mu+\mathrm{1}} \\ $$$$\theta\leqslant\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{1}\leqslant{m}\leqslant\mathrm{2}\mu+\mathrm{1} \\ $$$$ \\ $$$${x}_{{P}} =\frac{{a}}{\mathrm{tan}\:\theta}+{a}\:\mathrm{sin}\:\alpha={a}\left({m}+\mathrm{sin}\:\alpha\right) \\ $$$${y}_{{P}} ={a}−{a}\:\mathrm{cos}\:\alpha={a}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$${x}_{{Q}} ={b}−{b}\:\mathrm{cos}\:\beta={a}\mu\left(\mathrm{1}−\mathrm{cos}\:\beta\right) \\ $$$${y}_{{Q}} ={y}_{{B}} +{b}\:\mathrm{sin}\:\beta={a}\left[\mathrm{1}+\sqrt{\left(\mathrm{1}+{m}\right)\left(\mathrm{2}\mu+\mathrm{1}−{m}\right)}+\mu\:\mathrm{sin}\:\beta\right] \\ $$$$ \\ $$$${area}\:{of}\:{triangle}\:{OPQ}: \\ $$$$\Delta=\frac{{x}_{{Q}} {y}_{{Q}} }{\mathrm{2}}+\frac{\left({x}_{{P}} −{x}_{{Q}} \right)\left({y}_{{P}} +{y}_{{Q}} \right)}{\mathrm{2}}−\frac{{x}_{{P}} {y}_{{P}} }{\mathrm{2}} \\ $$$$\mathrm{2}\Delta={x}_{{P}} {y}_{{Q}} −{x}_{{Q}} {y}_{{P}} \\ $$$$\mathrm{2}\Delta={a}^{\mathrm{2}} \left({m}+\mathrm{sin}\:\alpha\right)\left(\mathrm{1}+\sqrt{\left(\mathrm{1}+{m}\right)\left(\mathrm{2}\mu+\mathrm{1}−{m}\right)}+\mu\:\mathrm{sin}\:\beta\right)−{a}^{\mathrm{2}} \mu\left(\mathrm{1}−\mathrm{cos}\:\beta\right)\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$$\frac{\mathrm{2}\Delta}{\mu{a}^{\mathrm{2}} }=\left({m}+\mathrm{sin}\:\alpha\right)\left(\frac{\mathrm{1}+\sqrt{\left(\mathrm{1}+{m}\right)\left(\mathrm{2}\mu+\mathrm{1}−{m}\right)}}{\mu}+\mathrm{sin}\:\beta\right)−\left(\mathrm{1}−\mathrm{cos}\:\beta\right)\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$${let}\:\Phi=\frac{\mathrm{2}\Delta}{\mu{a}^{\mathrm{2}} },\:\lambda=\frac{\mathrm{1}+\sqrt{\left(\mathrm{1}+{m}\right)\left(\mathrm{2}\mu+\mathrm{1}−{m}\right)}}{\mu} \\ $$$$\Phi=\left({m}+\mathrm{sin}\:\alpha\right)\left(\lambda+\mathrm{sin}\:\beta\right)−\left(\mathrm{1}−\mathrm{cos}\:\beta\right)\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$$\Phi={m}\lambda−\mathrm{1}+\lambda\:\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha+{m}\:\mathrm{sin}\:\beta+\mathrm{cos}\:\beta−\mathrm{cos}\:\left(\alpha+\beta\right) \\ $$$$\Phi={m}\lambda−\mathrm{1}+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{cos}\:\left(\alpha−\mathrm{tan}^{−\mathrm{1}} \lambda\right)+\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\:\mathrm{cos}\:\left(\beta−\mathrm{tan}^{−\mathrm{1}} {m}\right)−\mathrm{cos}\:\left(\alpha+\beta\right) \\ $$$$\Phi\:{is}\:{a}\:{function}\:{of}\:{three}\:{variables}:\: \\ $$$${m},\:\alpha,\:\beta. \\ $$$$ \\ $$$${i}'{ll}\:{try}\:{in}\:{geometric}\:{way}: \\ $$$${for}\:{maiximum}\:{area}\:{of}\:{triangle}\:{OPQ}, \\ $$$${tangent}\:{at}\:{Q}\:{must}\:{be}\://\:{OP}: \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{{y}_{{P}} }{{x}_{{P}} }=\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{{m}+\mathrm{sin}\:\alpha} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\beta}{\mathrm{sin}\:\beta}=\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{{m}+\mathrm{sin}\:\alpha} \\ $$$${m}\:\mathrm{cos}\:\beta−\mathrm{sin}\:\beta+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\mathrm{0} \\ $$$${m}\:\mathrm{cos}\:\beta−\mathrm{sin}\:\beta+\mathrm{sin}\:\left(\alpha+\beta\right)=\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\:\mathrm{sin}\:\left(\beta−\mathrm{tan}^{−\mathrm{1}} {m}\right)=\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$\left({we}\:{can}\:{get}\:{the}\:{same}\:{also}\:{from}\:\frac{\partial\Phi}{\partial\beta}=\mathrm{0}\right) \\ $$$$ \\ $$$${tangent}\:{at}\:{P}\:{must}\:{be}\://\:{OQ}: \\ $$$$\mathrm{tan}\:\alpha=\frac{{y}_{{Q}} }{{x}_{{Q}} }=\frac{\mathrm{1}+\sqrt{\left(\mathrm{1}+{m}\right)\left(\mathrm{2}\mu+\mathrm{1}−{m}\right)}+\mu\:\mathrm{sin}\:\beta}{\mu\left(\mathrm{1}−\mathrm{cos}\:\beta\right)} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}=\frac{\lambda+\mathrm{sin}\:\beta}{\mathrm{1}−\mathrm{cos}\:\beta} \\ $$$$\lambda\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta=\mathrm{0} \\ $$$$\lambda\:\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha+\mathrm{sin}\:\left(\alpha+\beta\right)=\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\left(\alpha−\mathrm{tan}^{−\mathrm{1}} \lambda\right)=\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$\left({we}\:{can}\:{get}\:{the}\:{same}\:{also}\:{from}\:\frac{\partial\Phi}{\partial\alpha}=\mathrm{0}\right) \\ $$$$ \\ $$$${we}\:{can}\:{see}\:{that}\:\frac{\pi}{\mathrm{2}}<\alpha+\beta<\pi\:{in}\:{case}\:{of} \\ $$$${that}\:\Delta\:{is}\:{maximum}. \\ $$$$ \\ $$$${say}\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\left(\alpha−\mathrm{tan}^{−\mathrm{1}} \lambda\right)=\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\:\mathrm{sin}\:\left(\beta−\mathrm{tan}^{−\mathrm{1}} {m}\right)=\mathrm{sin}\:\left(\alpha+\beta\right)=\xi \\ $$$$\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \lambda+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$$\Rightarrow\beta=\mathrm{tan}^{−\mathrm{1}} {m}+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$$\Rightarrow\alpha+\beta=\pi−\mathrm{sin}^{−\mathrm{1}} \xi \\ $$$$\begin{array}{|c|}{\mathrm{sin}^{−\mathrm{1}} \xi+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}=\pi−\mathrm{tan}^{−\mathrm{1}} \lambda−\mathrm{tan}^{−\mathrm{1}} {m}}\\\hline\end{array}\:\:\:…\left({I}\right) \\ $$$$ \\ $$$$\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\left(\alpha−\mathrm{tan}^{−\mathrm{1}} \lambda\right)=\xi \\ $$$$\Rightarrow\mathrm{cos}\:\left(\alpha−\mathrm{tan}^{−\mathrm{1}} \lambda\right)=\sqrt{\mathrm{1}−\frac{\xi^{\mathrm{2}} }{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$$\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\:\mathrm{sin}\:\left(\beta−\mathrm{tan}^{−\mathrm{1}} {m}\right)=\xi \\ $$$$\Rightarrow\mathrm{cos}\:\left(\beta−\mathrm{tan}^{−\mathrm{1}} {m}\right)=\sqrt{\mathrm{1}−\frac{\xi^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} }} \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)=\xi \\ $$$$\Rightarrow\mathrm{cos}\:\left(\alpha+\beta\right)=−\sqrt{\mathrm{1}−\xi^{\mathrm{2}} } \\ $$$$\Phi={m}\lambda−\mathrm{1}+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{cos}\:\left(\alpha−\mathrm{tan}^{−\mathrm{1}} \lambda\right)+\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\:\mathrm{cos}\:\left(\beta−\mathrm{tan}^{−\mathrm{1}} {m}\right)−\mathrm{cos}\:\left(\alpha+\beta\right) \\ $$$$\begin{array}{|c|}{\Phi={m}\lambda−\mathrm{1}+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} −\xi^{\mathrm{2}} }+\sqrt{\mathrm{1}+{m}^{\mathrm{2}} −\xi^{\mathrm{2}} }\:+\sqrt{\mathrm{1}−\xi^{\mathrm{2}} }}\\\hline\end{array}\:\:\:…\left({II}\right) \\ $$
Commented by mr W last updated on 05/Feb/22
Commented by mr W last updated on 05/Feb/22
Commented by mr W last updated on 05/Feb/22
Commented by mr W last updated on 05/Feb/22
$${the}\:{positions}\:{of}\:{the}\:{circles}\:{are}\:{defined} \\ $$$${by}\:{the}\:{parameter}\:{m}=\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\:{with} \\ $$$$\mathrm{1}\leqslant{m}\leqslant\mathrm{2}\mu+\mathrm{1}. \\ $$$${for}\:{a}\:{given}\:{m}\:{we}\:{can}\:{determine}\:{the} \\ $$$${parameter}\:\xi\:{which}\:{defines}\:{the}\: \\ $$$${positions}\:{of}\:{points}\:{P}\:{and}\:{Q}\:{on}\:{the} \\ $$$${circles},\:{i}.{e}.\:\alpha\:{and}\:\beta,\:{such}\:{that}\:\Delta_{{OPQ}} \:{is}\: \\ $$$${maximum}\:{relatively}\:{to}\:{the}\:{given}\:{m}. \\ $$$${but}\:{since}\:{the}\:{equation}\:{for}\:\xi\:{can}\:{not}\:{be} \\ $$$${solved}\:{analytically},\:{i}\:{have}\:{not}\:{found} \\ $$$${a}\:{way}\:{to}\:{determine}\:{the}\:{m}\:{for}\:{the} \\ $$$${absolute}\:{maximum}\:\Delta_{{OPQ}} . \\ $$
Commented by aleks041103 last updated on 05/Feb/22
$${In}\:{a}\:{very}\:{cute}\:{vectorial}\:{way}\:{you}\:{can} \\ $$$${prove}\:{that}\:{for}\:{max}\:{area} \\ $$$${OP}\bot{QB}\:{and}\:{OQ}\bot{PA} \\ $$$${It}\:{is}\:{possible}\:{to}\:{prove}\:{that}\:{for}\:{max}\:{area} \\ $$$$\frac{{x}_{{P}} }{{y}_{{Q}} }=\frac{{y}_{{B}} −{y}_{{A}} }{{x}_{{A}} −{x}_{{B}} } \\ $$$${later}\:{i}'{ll}\:{post}\:{a}\:{proof}!\:{I}\:{hope}\:{that}\:{helps}! \\ $$
Commented by mr W last updated on 05/Feb/22
$${that}'{s}\:{right}\:{sir}.\:{thanks}! \\ $$$${this}\:{is}\:{also}\:{that}\:{what}\:{my}\:{diagram}\: \\ $$$${at}\:{the}\:{beginning}\:{shows}: \\ $$$${tangent}\:{at}\:{Q}\://\:{AP},\:{i}.{e}.\:{BQ}\bot{OP} \\ $$$${tangent}\:{at}\:{P}\://\:{AQ},\:{i}.{e}.\:{AP}\bot{OQ} \\ $$
Commented by Tawa11 last updated on 05/Feb/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 06/Feb/22
$${Now}\:{i}\:{have}\:{found}\:{a}\:{way}\:{to}\:{find} \\ $$$${the}\:{absolute}\:{maximum}\:{area}\:{of}\:{the} \\ $$$${triangle}\:{and}\:{the}\:{corresponding}\: \\ $$$${parameter}\:{m}. \\ $$$${The}\:{relationship}\:{between}\:{the}\:{paramters} \\ $$$$\xi\:{and}\:{m}\:{is}\:{the}\:{equation}\:{I}: \\ $$$$\begin{array}{|c|}{\mathrm{sin}^{−\mathrm{1}} \xi+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}=\pi−\mathrm{tan}^{−\mathrm{1}} \lambda−\mathrm{tan}^{−\mathrm{1}} {m}}\\\hline\end{array} \\ $$$${the}\:{area}\:{of}\:{the}\:{triangle}\:{is}\:{expressed} \\ $$$${by}\:{the}\:{equation}\:{II}: \\ $$$$\begin{array}{|c|}{\Phi={m}\lambda−\mathrm{1}+\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} −\xi^{\mathrm{2}} }+\sqrt{\mathrm{1}+{m}^{\mathrm{2}} −\xi^{\mathrm{2}} }\:+\sqrt{\mathrm{1}−\xi^{\mathrm{2}} }}\\\hline\end{array} \\ $$$${with}\:\Phi=\frac{\mathrm{2}\Delta}{\mu{a}^{\mathrm{2}} }. \\ $$$${the}\:{maximum}\:\Phi\:{is}\:{when}\:{the}\:{curves} \\ $$$${of}\:{both}\:{equations}\:{just}\:{touch}\:{each} \\ $$$${other},\:{as}\:{following}\:{diagram}\:{shows}. \\ $$$$ \\ $$$$\underline{{example}:} \\ $$$$\mu=\frac{{b}}{{a}}=\mathrm{0}.\mathrm{75} \\ $$$$\Phi_{{max}} =\mathrm{11}.\mathrm{125757}\:{or}\: \\ $$$$\Delta_{{max}} =\frac{\mu{a}^{\mathrm{2}} \Phi_{{max}} }{\mathrm{2}}=\mathrm{4}.\mathrm{172}{a}^{\mathrm{2}} \\ $$$${at}\:{m}=\mathrm{1}.\mathrm{9858}\:{or}\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{m}}\right) \\ $$
Commented by mr W last updated on 06/Feb/22
Commented by mr W last updated on 06/Feb/22
Commented by mr W last updated on 06/Feb/22
Commented by mr W last updated on 06/Feb/22
Commented by mr W last updated on 06/Feb/22
Commented by mr W last updated on 06/Feb/22
Commented by mr W last updated on 06/Feb/22
Commented by mr W last updated on 06/Feb/22
$${further}\:{examples}: \\ $$$$\mu=\frac{{b}}{{a}}=\mathrm{0}.\mathrm{4} \\ $$$$\Phi_{{max}} =\mathrm{13}.\mathrm{51418}\:{or}\: \\ $$$$\Delta_{{max}} =\frac{\mu{a}^{\mathrm{2}} \Phi_{{max}} }{\mathrm{2}}=\mathrm{2}.\mathrm{703}{a}^{\mathrm{2}} \\ $$$${at}\:{m}=\mathrm{1}.\mathrm{3871} \\ $$