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2015-11-06T16:56:10+05:30

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Sum of the two positive numbers = 108
Let one number = x
other number = 108-x
given that the product of the first times the square of the second is a maximum or (108-x) × x²  is maximum.

(108-x) × x² = 108x² - x³

 \frac{d}{dx}(108 x^{2} - x^{3}  ) =0\\ \\ \Rightarrow 108 \times 2x - 3x^2 = 0\\ \\ \Rightarrow 216x - 3x^2 = 0\\ \\ \Rightarrow 3x^2 = 216x\\ \\ \Rightarrow 3x = 216\ \ \ \ or\ \ \ \ x = 0\\ \\ \Rightarrow x =  \frac{216}{3} =72\ \ \ \ \ or \ \ \ \ \ x=0

 \frac{d^2}{dx}(108x^2-x^3) =  \frac{d}{dx}(216x-3x^2)=216-6x\\ \\ for\ x=72,\ \   \frac{d^2}{dx}(108x^2-x^3) = 216-6 \times 72 = -216\ \textless \ 0\\ \\ So\ (108x^2-x^3)\ is\ maximum\ at\ x=72

So the two positive numbers are:
x = 72
108-x = 108 - 72 = 36
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