Word problem about exponential equation

Question

I have the following exponential equation problem:

A company launches a new product and has the following function describing the daily units of that product sold: $$ f_{k}(t)=k(t-15)e^{-0.001t}+15k $$ where t is the time since the release of the product (in days) and k is some constant. The company only makes a profit if it sells more than 4500 units daily. When does the company stop making profits?

What I have so far is the following:

\begin{align} k(t-15)e^{-0.001t}+15k&<4500 \tag 1\\ (t-15)e^{-0.001t}&<C \tag 2\\ \ln(t-15)-0.001t&<C' \tag 3 \end{align} where C is a constant.I don't know how to procede from here. The only thing I can think of is exponentiating both sides but then I would be just going back to the previous step.

Answer 1

You need a lambert W function. For $x \ge 0$ then $f(x) = x e^x$ is increasing and unbounded. So ever every $x \ge 0$ there is a unique $w$ so that $we^w = x$. So just as we define $\ln x= w$ to be the unique value where $e^w = x$. We define the lamber W function $W(x) = w$ to be the unique value where $w e^w = x$.

So $ k(t-15)e^{-0.001t}+15k <4500$

$(t-15)e^{-0.001t}< \frac {4500 -15k}k$ (I'm presuming $k > 0$)

$(t-15)e^{0.001t} > \frac k{4500-15k}$ (I'm presuming $15 < 4500$? And so we can deduce $t > 15$.)

$(t-15)e^{0.001(t-15)}e^{0.001\cdot 15} > \frac k{4500-15k}$

$0.001(t-15)e^{0.001(t-15)} > 0.001\frac k{4500-15k}e^{-0.001\cdot 15}=C_1$

$0.001(t-15) > W(C_1)$.

$t > 1000(W(C_1)) + 15$.

So it stops being profitable on day $t$ when $1000(W(C_1)) + 15 < t \le 1000(W(C_1)) + 15 + 1$ or when $t-1 \le 1000(W(C_1)) + 15< t$ or when $t- 1 =\lfloor 1000(W(C_1)) + 15\rfloor=\lfloor 1000(W(C_1)) \rfloor +15$ or when
$t = \lfloor 1000(W(C_1)) \rfloor + 16$.

There is no actual way to calculate $W(C_1)$ algebraically (just as there is no way to calculate $\ln K$ or $\sin x$ algebraically) but you can program or make tables for it.

Either you are either expected to know of the Lambert W function, or you know what $k$ is and are expected to plug in values to estimate.

It can't be done algebraically.

Answer 2

For $k<\frac{900 e^{203/200}}{200+3 e^{203/200}}\approx 11.924$, there is no time when the company will not make a profit.

This is a graph of $f_k(t)-4500$ for $k=11.9$ (blue) and $k=11.95$ (orange).

When there is a solution, the solution is $t=-5(-3+200 W((3((k-300) e^{3/200}))/(200 k)))$ where $W$ is the productlog function.
For $k$ large, the first order approximation $t=900000/(203k)$ works very well.
Here is a graph of the exact solution (blue) and the first order approximation (orange) for $k \in [100,200]$. They differ by about 2 days for $k=100$ and yby about 0.5 days for $k=200$.