Question Number 51869 by prakash jain last updated on 31/Dec/18
Commented by prakash jain last updated on 31/Dec/18
$$\mathrm{Correct}\:\mathrm{answer}\:\mathrm{given}\:\mathrm{is}\:\mathrm{2}/\mathrm{9}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{mininum}\:\mathrm{is}\:\mathrm{0}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18
![AB=1 BC=1 AC=(√2) let B (0,0) C(1,0) A(0,1) p(h,k) eqn of st line AC is x+y=1 h+k=1 [((h+k)/2)≥(√(hk)) (1/2)≥(√(hk)) so (1/4)≥hk] area of △APQ=(1/2)×h×k≤(1/2)×(1/4) area APQ≤(1/8) (area APQ)_(max) =(1/8) area PQBR=hk≤(1/4)(area PQBR_(max) =(1/4)) area PRC=areaABC−area APQ−areaPQBR =(1/2)×1×1−(1/2)×h×k−hk =(1/2)−(3/2)×hk now (1/2)−(3/2)×hk≥(1/2)−(3/2)×(1/4) ((1/2)−(3/2)hk)≥(1/8) max area APQ=(1/8) max area PQBR=(1/4) min area PRC.=(1/8) i could not understand the question however i have calculated the area pls check...](Q51886.png)
$${AB}=\mathrm{1}\:\:{BC}=\mathrm{1}\:\:{AC}=\sqrt{\mathrm{2}}\:\:\: \\ $$$${let}\:{B}\:\left(\mathrm{0},\mathrm{0}\right)\:\:\:{C}\left(\mathrm{1},\mathrm{0}\right)\:\:\:{A}\left(\mathrm{0},\mathrm{1}\right)\:\:{p}\left({h},{k}\right) \\ $$$${eqn}\:{of}\:{st}\:{line}\:{AC}\:{is}\:{x}+{y}=\mathrm{1} \\ $$$${h}+{k}=\mathrm{1}\:\:\left[\frac{{h}+{k}}{\mathrm{2}}\geqslant\sqrt{{hk}}\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\geqslant\sqrt{{hk}}\:\:\:\:{so}\:\frac{\mathrm{1}}{\mathrm{4}}\geqslant{hk}\right] \\ $$$${area}\:{of}\:\bigtriangleup{APQ}=\frac{\mathrm{1}}{\mathrm{2}}×{h}×{k}\leqslant\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:{area}\:{APQ}\leqslant\frac{\mathrm{1}}{\mathrm{8}}\:\:\left({area}\:{APQ}\right)_{{max}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${area}\:{PQBR}={hk}\leqslant\frac{\mathrm{1}}{\mathrm{4}}\left({area}\:{PQBR}_{{max}} =\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${area}\:{PRC}={areaABC}−{area}\:{APQ}−{areaPQBR} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}×{h}×{k}−{hk} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}×{hk} \\ $$$${now}\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}×{hk}\geqslant\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}{hk}\right)\geqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${max}\:{area}\:{APQ}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${max}\:{area}\:{PQBR}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${min}\:{area}\:{PRC}.=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{could}}\:\boldsymbol{{not}}\:\boldsymbol{{understand}}\:\boldsymbol{{the}}\:\boldsymbol{{question}} \\ $$$$\boldsymbol{{however}}\:\boldsymbol{{i}}\:\boldsymbol{{have}}\:\boldsymbol{{calculated}}\:\boldsymbol{{the}}\:\boldsymbol{{area}} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}}... \\ $$
Commented by prakash jain last updated on 01/Jan/19
Thanks. The question was asked in my son fiitjee pre NEW main test. I could not understand the question either.
Commented by peter frank last updated on 01/Jan/19
$${help}\:{me}\:{Qn}\:\mathrm{51901}\:{Qn}\mathrm{51817} \\ $$