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$$S=(150-3x)^{20}(2x-50)^{30}$$

What is the max value of S?

Given that 25< x < 50

Answer is $3^{50} \cdot 10^{50}$.

I need the explanation.

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  • $\begingroup$ Are you sure you copied it right? $\lim_{x\to\infty} S(x) = \lim_{x\to-\infty} S(x) = \infty$, so there is no maximum value. $\endgroup$ – JiK Feb 15 '14 at 15:12
  • $\begingroup$ correctd it,sorry $\endgroup$ – AJ_ Feb 15 '14 at 15:18
  • $\begingroup$ Still no difference. The absolute values of $150-3x$ and $2x-50$ will grow when $x$ grows, and the exponents are even so both $(150-3x)^{20}$ and $(2x-50)^{30}$ will be huge when $x$ is huge. $\endgroup$ – JiK Feb 15 '14 at 15:22
  • $\begingroup$ sorry again.now see :) $\endgroup$ – AJ_ Feb 15 '14 at 15:24
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After simplification, the derivative of $S(x)$ write $$60 (200-5 x) (150-3 x)^{19} (2 x-50)^{29}$$ So, it cancels for $x=40$ (the other roots would make $S=0$) and the maximum value is effectively $3^{50} \cdot 10^{50}$. You can check that this a maximum computing the value of the second derivative of $S$ with respect to $x$; it is given by $$k (x-50)^{18} (x-25)^{28} \left(49 x^2-3920 x+78250\right)$$ (I let you the pleasure of finding the value of the positive constant $k$); for $x=40$, its value is just $-3^{49} \cdot 10^{73}$

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You have to find the derivative and the roots of it.. Then you have to find the monotonicity of the function at each interval.

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