Question Number 15865 by prakash jain last updated on 14/Jun/17
$${a},{b},{c}\in\mathbb{R}^{+\:} \mathrm{andIf}\:{a}+{b}+{c}=\mathrm{18}\:\mathrm{then}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}^{\mathrm{4}} \:\mathrm{is} \\ $$
Answered by Tinkutara last updated on 14/Jun/17
![We know that A.M. ≥ G.M. ∴ (((a/2) + (a/2) + (b/3) + (b/3) + (b/3) + (c/4) + (c/4) + (c/4) + (c/4))/9) ≥ (((a^2 b^3 c^4 )/(2^2 .3^3 .4^4 )))^(1/9) ⇒ 2 ≥ (((a^2 b^3 c^4 )/(2^2 .3^3 .4^4 )))^(1/9) [∵ a + b + c = 18] ((a^2 b^3 c^4 )/(2^(10) .3^3 )) ≤ 2^9 ∴ a^2 b^3 c^4 ≤ 2^(19) .3^3](https://www.tinkutara.com/question/Q15870.png)
$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{A}.\mathrm{M}.\:\geqslant\:\mathrm{G}.\mathrm{M}. \\ $$$$\therefore\:\frac{\frac{{a}}{\mathrm{2}}\:+\:\frac{{a}}{\mathrm{2}}\:+\:\frac{{b}}{\mathrm{3}}\:+\:\frac{{b}}{\mathrm{3}}\:+\:\frac{{b}}{\mathrm{3}}\:+\:\frac{{c}}{\mathrm{4}}\:+\:\frac{{c}}{\mathrm{4}}\:+\:\frac{{c}}{\mathrm{4}}\:+\:\frac{{c}}{\mathrm{4}}}{\mathrm{9}}\:\geqslant\:\sqrt[{\mathrm{9}}]{\frac{{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{3}} .\mathrm{4}^{\mathrm{4}} }} \\ $$$$\Rightarrow\:\mathrm{2}\:\geqslant\:\sqrt[{\mathrm{9}}]{\frac{{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{3}} .\mathrm{4}^{\mathrm{4}} }}\:\left[\because\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{18}\right] \\ $$$$\frac{{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{10}} .\mathrm{3}^{\mathrm{3}} }\:\leqslant\:\mathrm{2}^{\mathrm{9}} \\ $$$$\therefore\:{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}^{\mathrm{4}} \:\leqslant\:\mathrm{2}^{\mathrm{19}} .\mathrm{3}^{\mathrm{3}} \\ $$
Answered by mrW1 last updated on 14/Jun/17
![c=18−(a+b) P=a^2 b^3 c^4 =a^2 b^3 (18−a−b)^4 (∂P/∂a)=2ab^3 (18−a−b)^4 −4a^2 b^3 (18−a−b)^3 =0 ⇒2ab^3 (18−a−b)^3 [(18−a−b)−2a]=0 ⇒(18−a−b)−2a=0 ...(i) (∂P/∂b)=3a^2 b^2 (18−a−b)^4 −4a^2 b^3 (18−a−b)^3 =0 (∂P/∂b)=3a^2 b^2 (18−a−b)^3 [(18−a−b)−(4/3)b]=0 ⇒(18−a−b)−(4/3)b=0 ...(ii) from (i) and (ii): 2a=(4/3)b a=(2/3)b 18−(2/3)b−b−(4/3)b=0 ⇒b=6 ⇒a=4 ⇒c=18−4−6=8 ⇒P_(max) =4^2 ×6^3 ×8^4 =2^(19) ×3^3 =14155776](https://www.tinkutara.com/question/Q15876.png)
$$\mathrm{c}=\mathrm{18}−\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{P}=\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{3}} \mathrm{c}^{\mathrm{4}} =\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{3}} \left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{4}} \\ $$$$ \\ $$$$\frac{\partial\mathrm{P}}{\partial\mathrm{a}}=\mathrm{2ab}^{\mathrm{3}} \left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{4}} −\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{3}} \left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2ab}^{\mathrm{3}} \left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{3}} \left[\left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)−\mathrm{2a}\right]=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)−\mathrm{2a}=\mathrm{0}\:\:\:…\left(\mathrm{i}\right) \\ $$$$ \\ $$$$\frac{\partial\mathrm{P}}{\partial\mathrm{b}}=\mathrm{3a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{4}} −\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{3}} \left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\frac{\partial\mathrm{P}}{\partial\mathrm{b}}=\mathrm{3a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{3}} \left[\left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{b}\right]=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{18}−\mathrm{a}−\mathrm{b}\right)−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{b}=\mathrm{0}\:\:\:\:…\left(\mathrm{ii}\right) \\ $$$$ \\ $$$$\mathrm{from}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right): \\ $$$$\mathrm{2a}=\frac{\mathrm{4}}{\mathrm{3}}\mathrm{b} \\ $$$$\mathrm{a}=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{b} \\ $$$$\mathrm{18}−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{b}−\mathrm{b}−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{b}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{6} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{c}=\mathrm{18}−\mathrm{4}−\mathrm{6}=\mathrm{8} \\ $$$$ \\ $$$$\Rightarrow\mathrm{P}_{\mathrm{max}} =\mathrm{4}^{\mathrm{2}} ×\mathrm{6}^{\mathrm{3}} ×\mathrm{8}^{\mathrm{4}} =\mathrm{2}^{\mathrm{19}} ×\mathrm{3}^{\mathrm{3}} =\mathrm{14155776} \\ $$