Question-46978

Question Number 46978 by rahul 19 last updated on 03/Nov/18

Answered by MrW3 last updated on 03/Nov/18

(q/p)=4−p⇒p=4−(q/p) (r/q)=4−q⇒q=4−(r/q) (p/r)=4−r⇒r=4−(p/r) ⇒p+q+r=12−((q/p)+(r/q)+(p/r))≤12−3×^3 (√((q/p)×(r/q)×(p/r)))=12−3=9 ⇒answer C) is correct

$$\frac{{q}}{{p}}=\mathrm{4}−{p}\Rightarrow{p}=\mathrm{4}−\frac{{q}}{{p}} \\ $$$$\frac{{r}}{{q}}=\mathrm{4}−{q}\Rightarrow{q}=\mathrm{4}−\frac{{r}}{{q}} \\ $$$$\frac{{p}}{{r}}=\mathrm{4}−{r}\Rightarrow{r}=\mathrm{4}−\frac{{p}}{{r}} \\ $$$$\Rightarrow{p}+{q}+{r}=\mathrm{12}−\left(\frac{{q}}{{p}}+\frac{{r}}{{q}}+\frac{{p}}{{r}}\right)\leqslant\mathrm{12}−\mathrm{3}×\:^{\mathrm{3}} \sqrt{\frac{{q}}{{p}}×\frac{{r}}{{q}}×\frac{{p}}{{r}}}=\mathrm{12}−\mathrm{3}=\mathrm{9} \\ $$$$ \\ $$$$\left.\Rightarrow{answer}\:{C}\right)\:{is}\:{correct} \\ $$

Commented by rahul 19 last updated on 04/Nov/18

thanks sir ��

Answered by ajfour last updated on 03/Nov/18

p+q+r=4(p+q+r)−(p^2 +q^2 +r^2 ) ⇒ p^2 +q^2 +r^2 =3(p+q+r) for maximum p+q+r p=q=r = s (say); then 3s^2 =9s ⇒ s = 0, 3 (p+q+r)_(max) = 3s = 9 .

$$\:\:\:{p}+{q}+{r}=\mathrm{4}\left({p}+{q}+{r}\right)−\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{3}\left({p}+{q}+{r}\right) \\ $$$$\:\:\:\:\:\:{for}\:{maximum}\:{p}+{q}+{r}\:\:\:\: \\ $$$${p}={q}={r}\:=\:{s}\:\:\left({say}\right);\:{then}\: \\ $$$$\:\:\mathrm{3}{s}^{\mathrm{2}} =\mathrm{9}{s}\:\:\Rightarrow\:\:{s}\:=\:\mathrm{0},\:\mathrm{3} \\ $$$$\left({p}+{q}+{r}\right)_{{max}} \:=\:\mathrm{3}{s}\:=\:\mathrm{9}\:. \\ $$

Commented by rahul 19 last updated on 04/Nov/18

thanks sir ��

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