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Find-the-minimum-value-of-5t-2-8t-5-2-3-t-2-2t-2-3-where-2-3-lt-t-lt-2-3-

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Question Number 197882 by CrispyXYZ last updated on 02/Oct/23
Find the minimum value of ((5t^2 −8t+5)/((2+(√3))t^2 −2t+2−(√3))) where 2−(√3)<t<2+(√3).
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{5}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}−\sqrt{\mathrm{3}}}\:\:\mathrm{where}\:\mathrm{2}−\sqrt{\mathrm{3}}<{t}<\mathrm{2}+\sqrt{\mathrm{3}}. \\ $$
Answered by mr W last updated on 03/Oct/23
((5t^2 −8t+5)/((2+(√3))t^2 −2t+2−(√3)))=(1/k), say (2+(√3)−5k)t^2 −2(1−4k)t+2−(√3)−5k=0 Δ=(1−4k)^2 −(2+(√3)−5k)(2−(√3)−5k)≥0 (1−4k)^2 −(2−5k)^2 +3≥0 k(3k−4)≤0 ⇒0≤k≤(4/3) ⇒(3/4)≤(1/k)<+∞ ⇒((5t^2 −8t+5)/((2+(√3))t^2 −2t+2−(√3)))≥(3/4)=minimun at t=((1−4k)/(2+(√3)−5k))=((1−((16)/3))/(2+(√3)−((20)/3)))=((14+3(√3))/(13)) 2−(√3)<((14+3(√3))/(13))<2+(√3)
$$\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{5}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}−\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{{k}},\:{say} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{5}{k}\right){t}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\mathrm{4}{k}\right){t}+\mathrm{2}−\sqrt{\mathrm{3}}−\mathrm{5}{k}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{1}−\mathrm{4}{k}\right)^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{5}{k}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}−\mathrm{5}{k}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{1}−\mathrm{4}{k}\right)^{\mathrm{2}} −\left(\mathrm{2}−\mathrm{5}{k}\right)^{\mathrm{2}} +\mathrm{3}\geqslant\mathrm{0} \\ $$$${k}\left(\mathrm{3}{k}−\mathrm{4}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant{k}\leqslant\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}\leqslant\frac{\mathrm{1}}{{k}}<+\infty \\ $$$$\Rightarrow\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{5}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}−\sqrt{\mathrm{3}}}\geqslant\frac{\mathrm{3}}{\mathrm{4}}={minimun} \\ $$$${at}\:{t}=\frac{\mathrm{1}−\mathrm{4}{k}}{\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{5}{k}}=\frac{\mathrm{1}−\frac{\mathrm{16}}{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}−\frac{\mathrm{20}}{\mathrm{3}}}=\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}} \\ $$$$\mathrm{2}−\sqrt{\mathrm{3}}<\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}}<\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

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