# Thread: find derivative, set to zero (partially solved...maybe)

1. ## find derivative, set to zero (partially solved...maybe)

I need to find the maximum of this equation by finding the derivative and setting it to zero:

$f(x) = (1+20\sqrt{t})e^{-.05t} = e^{-.05t} + 20\sqrt{t}e^{-.05t}$

Now, solving for the derivative using the chain rule for the first operand, and using product rule+chain rule for the second operand, I get the following:

$f'(x) = \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05}{e^{.05t}} = 0$

get common denominators:

$= \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05\sqrt{t}}{\sqrt{t}e^{.05t}}$

Assuming that is correct so far, where do I go from there? Do I multiply through to cancel the denominator to get the following:

$10 - t - .05\sqrt{t} = 0$

If that's correct, then what do I do?

2. Originally Posted by sgcb
I need to find the maximum of this equation by finding the derivative and setting it to zero:

$f(x) = (1+20\sqrt{t})e^{-.05t} = e^{-.05t} + 20\sqrt{t}e^{-.05t}$

Now, solving for the derivative using the chain rule for the first operand, and using product rule+chain rule for the second operand, I get the following:

$f'(x) = \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05}{e^{.05t}} = 0$

get common denominators:

$= \frac{10-t}{\sqrt{t}e^{.05t}} - \frac{.05\sqrt{t}}{\sqrt{t}e^{.05t}}$

Assuming that is correct so far, where do I go from there? Do I multiply through to cancel the denominator to get the following:

$10 - t - .05\sqrt{t} = 0$

If that's correct, then what do I do?
let $u = \sqrt{t}$ ...

$10 - u^2 - .05u = 0$

$u^2 + .05u - 10 = 0$

$u = \frac{-.05 \pm \sqrt{40.0025}}{2}$

since $\sqrt{t} > 0$ ...

$\sqrt{t} = \frac{-.05 + \sqrt{40.0025}}{2}$

$t \approx 9.843$

max $f(t) \approx 38.969$

3. Oh, cool! It never occurred to me to set it up that way, thanks skeeter! That was also the answer I got on the calculator.