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a-b-c-R-andIf-a-b-c-18-then-maximum-value-of-a-2-b-3-c-4-is-

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Question Number 15865 by prakash jain last updated on 14/Jun/17
a,b,c∈R^(+ ) andIf a+b+c=18 then maximum value of a^2 b^3 c^4 is
a,b,c∈R+andIfa+b+c=18thenmaximumvalueofa2b3c4is
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
WeknowthatA.M.⩾G.M.∴a2+a2+b3+b3+b3+c4+c4+c4+c49⩾a2b3c422.33.449⇒2⩾a2b3c422.33.449[∵a+b+c=18]a2b3c4210.33⩽29∴a2b3c4⩽219.33
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
c=18−(a+b)P=a2b3c4=a2b3(18−a−b)4∂P∂a=2ab3(18−a−b)4−4a2b3(18−a−b)3=0⇒2ab3(18−a−b)3[(18−a−b)−2a]=0⇒(18−a−b)−2a=0…(i)∂P∂b=3a2b2(18−a−b)4−4a2b3(18−a−b)3=0∂P∂b=3a2b2(18−a−b)3[(18−a−b)−43b]=0⇒(18−a−b)−43b=0…(ii)from(i)and(ii):2a=43ba=23b18−23b−b−43b=0⇒b=6⇒a=4⇒c=18−4−6=8⇒Pmax=42×63×84=219×33=14155776

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