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Question Number 131965 by liberty last updated on 10/Feb/21

Commented by liberty last updated on 10/Feb/21

Commented by TheSupreme last updated on 10/Feb/21

$$T=t_1+t_2$$$$1)\sqrt{3^2+x^2}=6t_1$$$$2)8-x=8t_2$$$$T=\frac{\sqrt{9+x^2}}{6}+1-\frac{x}{8}$$$$0<x<8$$$$\frac{dT}{dx}=\frac{x}{6\sqrt{9+x^2}}-\frac{1}{8}=0$$$$8x=6\sqrt{9+x^2}$$$$64x^2=36\ast 9+36x^2$$$$28x^2=36\ast 9$$$$x=\frac{9}{\sqrt{7}}$$$$T\left(0\right)=\frac{3}{2}$$$$T\left(8\right)=\frac{\sqrt{73}}{6}$$$$T\left(\frac{9}{\sqrt{7}}\right)=\frac{\sqrt{9+\frac{81}{7}}}{6}+1-\frac{9}{8\sqrt{7}}=\frac{\sqrt{\frac{144}{7}}}{6}+1-\frac{9}{8\sqrt{7}}=\sqrt{7}$$$$T_{min}=min(\frac{3}{2},\frac{\sqrt{7}}{8}+1,\frac{\sqrt{73}}{6})=\frac{\sqrt{7}}{8}+1$$$$sox=\frac{9}{\sqrt{7}}$$$$$$$$$$

Commented by liberty last updated on 10/Feb/21

$$\mathrm{yes}..=$$

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