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Prove that maximum value of a^2*b^3*c^4 subject to a+b+c=18 is 4^2*6^3*8^4. without using tat AM,GM equality thing...

Prove that maximum value of a^2*b^3*c^4 subject to a+b+c=18 is 4^2*6^3*8^4. without using tat AM,GM equality thing...

Grade:10

3 Answers

Chetan Mandayam Nayakar
312 Points
12 years ago

You can use the method of Lagrange Multiplifiers. It is not part of IITJEE syllabus(I am not fully sure regarding this). It is easy if you use this method.

Sathya
35 Points
12 years ago

Can u pls say  what r  Lagrange Multiplifiers  in detail?

Chetan Mandayam Nayakar
312 Points
12 years ago

Lagrange multipliers, also called Lagrangian multipliers (e.g., Arfken 1985, p. 945), can be used to find the of a multivariate function f(x_1,x_2,...,x_n) subject to the constraint g(x_1,x_2,...,x_n)=0, where f and g are functions with continuous first partial derivatives on the open set containing the curve g(x_1,x_2,...,x_n)=0, and del g!=0 at any point on the curve (where del is the gradient).

LagrangeMultipliers

For an extremum of f to exist on g, the gradient of f must line up with the gradient of g. In the illustration above, f is shown in red, g in blue, and the intersection of f and g is indicated in light blue. The gradient is a horizontal vector (i.e., it has no z-component) that shows the direction that the function increases; for g it is perpendicular to the curve, which is a straight line in this case. If the two gradients are in the same direction, then one is a multiple (-lambda) of the other, so

del f=-lambdadel g.
(1)

The two vectors are equal, so all of their components are as well, giving

(partialf)/(partialx_k)+lambda(partialg)/(partialx_k)=0
(2)

for all k=1, ..., n, where the constant lambda is called the Lagrange multiplier.

The extremum is then found by solving the n+1 equations in n+1 unknowns, which is done without inverting g, which is why Lagrange multipliers can be so useful.

For multiple constraints g_1=0, g_2=0, ...,

del f+lambda_1del g_1+lambda_2del g_2+...=0.

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