Q.113. Maximum Value, Application of Differentiation.

Question 113.
If a + b + c = 1 and a, b, c > 0, then find the maximum value of a*b^2*c^3.

Answer 113.
a + b + c = 1
=> a = 1 - b - c
=> f (b, c)
= (1 - b - c) * b^2c^3
= b^2c^3 - b^3c^3 - b^2c^4
 For f (b, c) to be maximum,
∂f/∂b = 2bc^3 - 3b^2c^3 - 2bc^4 = 0 and
∂f/∂c = 3b^2c^2 - 3b^3c^2 - 4b^2c^3 = 0
=> bc^3 (2 - 3b - 2c) = 0
and b^2c^2(3 - 3b - 4c) = 0

b = 0 or c = 0 gives the minimum bvalue of f (b, c).

Solving
2 - 3b - 2c = 0 and 3 - 3b - 4c = 0,
b = 1/3 and c = 1/2
=> a = 1 - 1/3 - 1/2 = 1/6
=> maximum value of a*b^2*c^3
= (1/6) *(1/3)^2 *(1/2)^3
= 1/(432).

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