The struggle is real! Find the time t in years when the annual sales x of a new product are increasing at the greatest rate. Use a graphing utility to verify your results.

Question

dan815 · Answer

whats the function

anonymous · Answer

it keeps kicking me out of the problem =/

anonymous · Answer

$$\frac{ 900000t^2 }{ 144+t^2 }$$

anonymous · Answer

@dan815

anonymous · Answer

So far I have (144+ t^2)(1800000t)-(900000t^2)(2t)

anonymous · Answer

hello!

anonymous · Answer

When they say graphing utility, do they want you to plug  it into the graphing calculator to find the largest slope?

anonymous · Answer

because "increasing at the greatest rate" part

anonymous · Answer

I think so, but I get error when I plug it in ,because its such a big number

anonymous · Answer

hmm do you have a ti81?

anonymous · Answer

ti-84 plus

anonymous · Answer

oh :/ you might just plug in manually with mathematics and then just drawing it out and estimate

aum · Answer

Factor 10^5 out and keep it as a constant outside.

anonymous · Answer

you just need values I think to justify that it's increasing at the greatest rate so, maybe. But yeah @aum that seems to be the best choice.

aum · Answer

If you want to solve this mathematically first and later verify your answer with the graphing calculator, then:
The first derivative gives the rate of increase.  For it to be a maximum, take the second derivative, equate it to zero and solve for t.

anonymous · Answer

I'm going to try that next 
I have the problem I believe to to the equate zero part and solve for t and got 3sqrt2, but webassign marked it wrong .

anonymous · Answer

err I got to the equation zero part
I think I did probably something wrong 
at 259200000(144+t^2)^2-(4t^3+576t)(259200000t)

anonymous · Answer

I know for sure though the bottom is (144+t^2)^3

aum · Answer

$$
f(t) = \frac{ 900000t^2 }{ 144+t^2 } = 10^5 *  \frac{ 9t^2 }{ 144+t^2 } \
f'(t) = 10^5 * \frac{ (144+t^2)*18t-9t^2(2t)  }{ (144+t^2)^2 } = 
 10^5 * \frac{ 2592t + 18t^3 - 18t^3 }{ (144+t^2)^2 } = \ 
f'(t) = 10^5 * \frac{ 2592t }{ (144+t^2)^2 } \
$$

aum · Answer

$$
f'(t) = 10^5 * \frac{ 2592t }{ (144+t^2)^2 } \
f''(t) = 10^5 * \frac{(144+t^2)^2(2592) - 2592t(2)(144+t^2)(2t)}{(144+t^2)^4}  = 0\
$$

aum · Answer

$$
2592(144+t^2)(144+t^2 - 4t^2) = 0 \
2592(144+t^2)(144- 3t^2) = 0 \
144 = 3t^2 \
t^2 = 48 = 16 * 3 \
t = 4\sqrt{3} \
$$

anonymous · Answer

WOW thank you so much!

aum · Answer

You are welcome.

anonymous · Answer

I have been struggling with this problem for the last two hours

aum · Answer

Glad to be able to help.

anonymous · Answer

Took a screen shot for future use:) Thanks again:)