Question Number 143858 by gsk2684 last updated on 19/Jun/21
$$\mathrm{if}\:\mathrm{A}\geqslant\mathrm{0},\:\mathrm{B}\geqslant\mathrm{0},\:\mathrm{A}+\mathrm{B}=\frac{\Pi}{\mathrm{3}}\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\: \\ $$$$\mathrm{of}\:\mathrm{tan}\:\mathrm{A}.\mathrm{tan}\:\mathrm{B}\: \\ $$
Answered by ajfour last updated on 19/Jun/21
$$\sqrt{\mathrm{3}}=\frac{{p}+{q}}{\mathrm{1}−{pq}} \\ $$$$\Rightarrow\:\mathrm{1}−{pq}=\frac{{p}+{q}}{\:\sqrt{\mathrm{3}}} \\ $$$${pq}=\mathrm{1}−\left(\frac{{p}+{q}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${if}\:{p}={q} \\ $$$${p}^{\mathrm{2}} +\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}=\mathrm{0} \\ $$$${p}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${p}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:\Rightarrow\:\:\left({pq}\right)_{{max}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left({pq}\right)_{{min}} =\mathrm{0} \\ $$
Commented by ajfour last updated on 19/Jun/21
$${A}\:{rectangle}\:{of}\:{given}\:{perimeter} \\ $$$${has}\:{maximum}\:{area}\:{when} \\ $$$${it}\:{is}\:{a}\:{square}. \\ $$
Commented by gsk2684 last updated on 19/Jun/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{have}\:\mathrm{a}\:\mathrm{doubt}\: \\ $$$$\mathrm{that}\:\mathrm{why}\:\mathrm{did}\:\mathrm{you}\:\mathrm{take}\:\mathrm{p}=\mathrm{q} \\ $$
Commented by gsk2684 last updated on 21/Jun/21
$$\mathrm{thanks}\:\mathrm{alot}. \\ $$