How Do You Find the Maximum of C(T) = 2t/(t+3)^2?

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can someone please help me with this, I've been having trouble with it.

C(T)= 2t/(t+3)^2

find the maximum?

thanks
 
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Do you know how to find the derivative of the function? If so then check when the slope is equal to 0.
To find the derivative use the (f'(x)g(x) - f(x)g'(x))/(g(x)^2)
 
If you've "been having trouble with it", then you must have tried something! What have you tried? Exactly where are you having trouble?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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