Question

find two positive rel numberssuch that they sum to 108 and the product of the first times the square of the second is a maximum

Answer 1

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

Answer 2

Answer:

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