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Question Number 165965 by ajfour last updated on 10/Feb/22

Answered by mr W last updated on 10/Feb/22

Commented by mr W last updated on 12/Feb/22

$$A=A_1+A_2$$$$A_{1}ismaximumwhenp=q=\frac{x}{\sqrt{2}}.$$$$A_{1,max}=\frac{x^2}{4}$$$$A_{2}ismaximumwhenthequadrilateral$$$$iscyclic.$$$$A_{2,max}=\sqrt{(s-a)(s-b)(s-c)(s-x)}$$$$withs=\frac{a+b+c+x}{2}.$$$$R=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$$$withd=x.$$$$A=\frac{x^2}{4}+\sqrt{(s-a)(s-b)(s-c)(s-x)}$$$$suchthatAismaximum,$$$$\frac{dA}{dx}=\frac{x}{2}+\frac{\sqrt{(s-a)(s-b)(s-c)(s-x)}}{4}(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}-\frac{1}{s-x})=0$$$$\Rightarrow 2x+\sqrt{(s-a)(s-b)(s-c)(s-x)}(\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}-\frac{1}{s-x})=0$$$$....exactsolutionforxseems$$$$impossible...$$$$$$$$example:a=1,b=3,c=2$$$$\Rightarrow x\approx 5.4969,A_{max}\approx 11.062$$

Commented by mr W last updated on 11/Feb/22

Commented by mr W last updated on 11/Feb/22

Commented by mr W last updated on 11/Feb/22

Commented by mr W last updated on 11/Feb/22

Commented by Tawa11 last updated on 11/Feb/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by aleks041103 last updated on 12/Feb/22

$$Theapproachyoutakeupisthesame$$$$towhatIhadinmind,butstillIhave$$$$anobjection.$$$$Youmaximized{A}_{1}foragivenx,and$$$$thenyouseemtoimplythatif{A}_{2}ismax$$$$thenA=A_1+A_{2}willbemax,which$$$${isn}^{\prime }tnessecairlytrue.$$$$Basically,yousaythat:$$$$A\left(x\right)=A_{1max}\left(x\right)+A_{2max}\left(x\right)=\frac{x^2}{4}+A_{2max}\left(x\right)$$$$ismaxwhen{A}_{2max}\left(x\right)ismax.$$$$Thisisobviouslynottrue,since:$$$$Aismax\Rightarrow A^{\prime }=0$$$$but$$$$A^{\prime }=\frac{x}{2}+A_{2max}^{\prime }=0\nLeftrightarrow A_{2max}^{\prime }=0$$

Commented by mr W last updated on 12/Feb/22

$$i{didn}^{\prime }tsaythatAismaximumif{A}_2$$$$ismaximum.youcanseethati$$$$setA=\frac{x^2}{4}+\sqrt{(s-a)(s-b)(s-c)(s-x)}$$$$andrequest\frac{dA}{dx}=0.$$$$whenwehavefoursidesa,b,c,x,{it}^{\prime }s$$$$clearthat{A}_{1}shouldbeanisosceles$$$$righttrianglewitharea\frac{x^2}{4}andthe$$$$quatrilateralshouldbeacyclicone$$$$witharea\sqrt{(s-a)(s-b)(s-c)(s-x)}.$$$$thesumofbothshouldbemaximum.$$

Commented by mr W last updated on 12/Feb/22

$$thesolutioncouldbemucheasierif$$$$wejustrequestthat{A}_{2}shouldbe$$$$maximum,seeQ166007.$$

Commented by aleks041103 last updated on 12/Feb/22

$$Yes,{you}^{\prime }reright!I^{\prime }msorry,I$$$$misunderstoodyoursolution!$$$${It}^{\prime }sfine!Sorryagain!$$

Commented by mr W last updated on 12/Feb/22

$${it}^{\prime }salrightsir.iappreciatesuch$$$$substantialfeedbackverymuch.$$$$thanksalot!$$

Commented by mr W last updated on 12/Feb/22

Commented by mr W last updated on 12/Feb/22

$$thisisthemaximalpossiblearea.$$$$buticannotexactlycalculateR.$$$$igetafinalequationwith{R}^6,thus$$$$unsolvable.$$

Commented by mr W last updated on 12/Feb/22

$$R=\sqrt{\frac{(ab+cx)(ac+bx)(bc+ax)}{(-a+b+c+x)(a-b+c+x)(a+b-c+x)(a+b+c-x)}}$$$$R=\frac{x}{\sqrt{2}}$$$$\frac{x^2}{2}=\frac{(ab+cx)(ac+bx)(bc+ax)}{(-a+b+c+x)(a-b+c+x)(a+b-c+x)(a+b+c-x)}$$$$\Rightarrow x^6-2(a^2+b^2+c^2)x^4-6{abcx}^3+(a^4+b^4+c^4)x^2+2abc(a^2+b^2+c^2)x+2a^2{b}^2{c}^2=0$$$$exactsolutionnotpossible...$$

Answered by ajfour last updated on 11/Feb/22

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