Author Topic: How would I go about this question? (Read 1216 times) Tweet Share

RuiAce · « **Reply #1 on:** May 24, 2017, 08:44:28 pm »

$\text{Simply substitute }t=0\text{ for the midday to get}\\ \frac{dE}{dt}=235$
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$\text{Since we want to maximise the rate, we will need to consider the second derivative}\\ \frac{dE}{dt}=220te^{-0.4t}+15 \implies \frac{d^2E}{dt^2}=220e^{-0.4t}-88te^{-0.4t}$
$\text{Set }\frac{d^2E}{dt^2}=0\text{ to maximise the rate}\\ \begin{align*}220e^{-0.4t}-88te^{-0.4t}&=0\\ e^{-0.4t}(220-88t)&=0\\ 220-88t&=0\\ t&=0.4\end{align*}$

HotDiggity · « **Reply #2 on:** May 25, 2017, 10:06:35 am »

Quote from: RuiAce on May 24, 2017, 08:44:28 pm

$\text{Simply substitute }t=0\text{ for the midday to get}\\ \frac{dE}{dt}=235$
______________
$\text{Since we want to maximise the rate, we will need to consider the second derivative}\\ \frac{dE}{dt}=220te^{-0.4t}+15 \implies \frac{d^2E}{dt^2}=220e^{-0.4t}-88te^{-0.4t}$
$\text{Set }\frac{d^2E}{dt^2}=0\text{ to maximise the rate}\\ \begin{align*}220e^{-0.4t}-88te^{-0.4t}&=0\\ e^{-0.4t}(220-88t)&=0\\ 220-88t&=0\\ t&=0.4\end{align*}$

if you subbed t=0 would it not be 15 is the rate then?

RuiAce · « **Reply #3 on:** May 25, 2017, 10:07:21 am »

I think I forgot about the t in the first term at the time. Yes it is 15

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