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14 Aug 2009 18:49:29 IST

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If a, b, c are +ve nos. such that 3a+2b+c=1, the greatest value of a^3 * b^3 * c^2 is?

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If a, b, c are +ve nos. such that 3a+2b+c=1, the greatest value of a^3 * b^3 * c^2 is?

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Yagyadutt Mishra

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14 Aug 2009 19:03:04 IST

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Yagyadutt Mishra

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14 Aug 2009 19:03:44 IST

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question is a^3.b^2.c..not a^3.b^3.c^2

®µD®A

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15 Aug 2009 10:24:54 IST

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bladeX -rise of a new sun

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15 Aug 2009 10:49:37 IST

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Here I will use LaGrange's multiplier :

F(a,b,c) = 3a + 2b + c =1 is a constraint

G(a,b,c) = a^3 *b^3*c^2

L(a,b,c,l) = G(a,b,c) - l (F(a,b,c)-1) ..........l is the LaGrange's multiplier

Now partially differentiate L(a,b,c,l) 4 times with respect to a,b,c and l . And then we will get 4 eqn and just solve them to get a,b,c and put them in G(a,b,c).

For the contemporary +2 proof my friend has already given it before for which i have rated him/.

Bipin Dubey

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15 Aug 2009 10:50:25 IST

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Mr. Mishra, it can be done for a³b³c² also.

Apply AM >= GM for the numbers : a, a, a, 2b/3, 2b/3, 2b/3, c/2, c/2

$\frac{a+a+a+\frac{2b}{3}+\frac{2b}{3}+\frac{2b}{3}+\frac{c}{2}+\frac{c}{2}}{8}\geq \left( (a)(a)(a)\left( \frac{2b}{3}\right)\left( \frac{2b}{3}\right)\left( \frac{2b}{3}\right)\left( \frac{c}{2}\right)\left( \frac{c}{2}\right)\right)^{1/8}$

$\left( \frac{2a^{3}b^{3}c^{2}}{27}\right)^{1/8}\leq \frac{3a+2b+c}{8}$

$\left( \frac{2a^{3}b^{3}c^{2}}{27}\right)^{1/8}\leq \frac{1}{8}$

$a^{3}b^{3}c^{2}\leq \left( \frac{27}{2}\right)\left( \frac{1}{8}\right)^{8}$

bladeX -rise of a new sun

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15 Aug 2009 10:51:07 IST

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For those having little knowledge of LaGrange's multiplier read the article in wikipedia and watch the lectures on youtube in NPTEL. i did the same

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