Question Number 164437 by SLVR last updated on 17/Jan/22
$${Find}\:{the}\:{max}.\:{area}\:{of}\:\Delta{le} \\ $$ $${witth}\:{sides}\:{a},{b},{c}\:{such}\:{as} \\ $$ $$\mathrm{0}<{a}\leqslant\mathrm{1};\mathrm{1}\leqslant{b}\leqslant\mathrm{2};\mathrm{2}\leqslant{c}\leqslant\mathrm{3}\:{is} \\ $$ $$\left.{a}\left.\right)\left.\mathrm{1}\left.\:\:\:\:\:\:\:\:{b}\right)\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:{c}\right)\mathrm{2}\:\:\:\:\:\:\:\:{d}\right)\mathrm{3}/\mathrm{2} \\ $$
Answered by mahdipoor last updated on 17/Jan/22
$${get}\:{a}\:{and}\:{b}\:{are}\:{constad}\:\Rightarrow \\ $$ $${S}=\frac{{ab}}{\mathrm{2}}{sin}\theta\:\Rightarrow\frac{{dS}}{{d}\theta}=\mathrm{0}=\frac{{ab}}{\mathrm{2}}{cos}\theta\:\Rightarrow\theta=\mathrm{90}^{°} \\ $$ $$\Rightarrow\:{a}\:,\:{b}\:,\:{c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:}\:\:\:{S}=\frac{{ab}}{\mathrm{2}} \\ $$ $${for}\:{a}=\mathrm{1}\:{and}\:{b}=\mathrm{2}\:\:\left({and}\:{c}=\sqrt{\mathrm{5}}\right) \\ $$ $$\Rightarrow{S}_{{max}} =\frac{\mathrm{1}×\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$
Commented bySLVR last updated on 17/Jan/22
$${extreemly}\:{great}\:{sir}...{god}\:{bless} \\ $$ $${you}\:{sir} \\ $$