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my math book gives the following question: A company sells phones and models the daily sales with the following function: $$f(t) = k*(t-15)*e^{-0,01t}+k*15$$

I have to find the value for t, so that f(t) = 4500 with k = 200

This is where I am stuck:$$200t*e^{-0,01t}-3000e+3000=4500$$

You get this equitation when you set 200 for k. I can't figure out how I have to transform the term to get the value of t.

Thanks!

Bobface
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2 Answers2

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When you see the variable you are trying to solve for in both the coefficient and exponent, the Lambert W function is always an appropriate path to go.

First, lets isolate the first term.

$200te^{-0.01t}=1500+3000e$

Now transform the first term so that the coefficient is the same as the exponent.

$te^{-0.01t}=7.5+15e$

$-0.01te^{-0.01t}=-0.075-0.15e$

Now take the "$W$" as follows:

$-0.01t=W(-0.075-0.15e)$

$t=-100W(-0.075-0.15e)$

Now, if you want to find the value of that as a decimal, look up "Lamber W calculator".

Simply Beautiful Art
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A less high-powered approach. I am assuming that since this is a model of daily sales that $t$ is the number of days, and it doesn't make sense to push the value of $t$ any further than the nearest integer. So we just want to find the day that sales pass 4500.

You went a step too far already. you want to put it in this form: $$200(t -15)e^{-0.01t} + 3000 = 4500$$ Then solve it for the $t$ not in the exponent: $$200(t -15)e^{-0.01t} = 1500$$ $$(t - 15)e^{-0.01t} = 7.5$$ $$t -15 = 7.5e^{0.01t}$$ $$t = 15 + 7.5e^{t/100}$$ Since $t$ is positive, and is evidently much less than $100$, we can expect that $1<e^{t/100}<2$.

Now we go through a process of refining that estimate by use the equation above to produce new limits for $t$. Those new limits provide new limits for $e^{t/100}$, which provides new limits for $t$, etc. We keep this up until the limits on $t$ are within a single day: $$1<e^{t/100}<2$$ $$22.5 = 15 + 7.5\cdot 1 < t < 15 + 7.5\cdot 2 =30 $$ $$1.25 \approx e^{0.225 }< e^{t/100} < e^{0.3} \approx 1.35$$ $$24.39 < t < 25.124$$ $$1.276 \approx e^{0.2439} < e^{t/100} < e^{0.25124} \approx 1.286$$ $$24.57 < t < 24.64$$

So by the model, sales reach 4500 on day 24.

Paul Sinclair
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