Question Number 36590 by ajfour last updated on 03/Jun/18
Commented by ajfour last updated on 03/Jun/18
$${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{ABC} \\ $$$${if}\:{one}\:{vertex}\:{is}\:{on}\:{smaller}\:{circle} \\ $$$${of}\:{radius}\:\boldsymbol{{r}}\:{and}\:{other}\:{two}\:{on}\: \\ $$$${larger}\:{circle}\:{of}\:{radius}\:\boldsymbol{{R}}\:. \\ $$$${The}\:{distance}\:{between}\:{centres}\:{of} \\ $$$${circles}\:{being}\:\boldsymbol{{d}}. \\ $$
Answered by ajfour last updated on 03/Jun/18
![△=(1/2)(x+d+r)(√(R^2 −x^2 )) (d△/dx)=((√(R^2 −x^2 ))/2)−((x(x+d+r))/(2(√(R^2 −x^2 )))) =0 ⇒ R^2 −x^2 =x^2 +x(d+r) 2x^2 +(d+r)x−R^2 =0 x=(((√((d+r)^2 +8R^2 ))−(d+r))/4) △_(max) =(1/2)[(((√((d+r)^2 +8R^2 ))+3(d+r))/4)] ×(√(R^2 −((((√((d+r)^2 +8R^2 ))−(d+r))/4))^2 )) .](https://www.tinkutara.com/question/Q36592.png)
$$\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{d}+{r}\right)\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\frac{{d}\bigtriangleup}{{dx}}=\frac{\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{2}}−\frac{{x}\left({x}+{d}+{r}\right)}{\mathrm{2}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:{R}^{\mathrm{2}} −{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +{x}\left({d}+{r}\right) \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\left({d}+{r}\right){x}−{R}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{\sqrt{\left({d}+{r}\right)^{\mathrm{2}} +\mathrm{8}{R}^{\mathrm{2}} }−\left({d}+{r}\right)}{\mathrm{4}} \\ $$$$\bigtriangleup_{{max}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\sqrt{\left({d}+{r}\right)^{\mathrm{2}} +\mathrm{8}{R}^{\mathrm{2}} }+\mathrm{3}\left({d}+{r}\right)}{\mathrm{4}}\right] \\ $$$$ \\ $$$$\:\:\:\:\:\:\:×\sqrt{{R}^{\mathrm{2}} −\left(\frac{\sqrt{\left({d}+{r}\right)^{\mathrm{2}} +\mathrm{8}{R}^{\mathrm{2}} }−\left({d}+{r}\right)}{\mathrm{4}}\right)^{\mathrm{2}} }\:. \\ $$
Commented by MrW3 last updated on 11/Jun/18
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