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The production of a mine is decreasing exponentially,and in the past $5$ years there has been a decline of $18\%$.If production declines by $90\%$,the mine will close.

The equation of production $P$ after $t$ years is given by $P=500+6500e^{-kt}$. Find

(a) the percentage of production decline after $10$ years.

(b) how long it will take for the mine to close.

I am stuck on this exponential decay question. I'v tried to find what $k$ is and substitute this value in to the formula with $t= 10$ years to find $P$ with no success.

I keep getting a negative answer.

I need some help with the working.

nmasanta
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user694587
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1 Answers1

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It is given that

$\displaystyle P(t) =500+6500e^{-kt}$

Initially when $t = 0$, we have $\displaystyle P(0) = 500 + 6500 = 7000$

After $5$ years, $\displaystyle P(5) = 500+6500e^{-5k} = (1-0.18)*P(0) = 0.82*7000 = 5740$

or, $\displaystyle 6500e^{-5k} = 5740 -500 = 5240$

or, $\displaystyle e^{-5k} = 5740 -500 = \frac{5240}{6500}$

or, $\displaystyle k = \frac{\ln \left(\frac{5240}{6500} \right)}{-5} = 0.04309613571372812633 \approx 0.0431$

(a) After $10$ years

$\displaystyle P(10) = 500+6500e^{-10k} = 4724.2462$

Percentage decline

$\displaystyle \frac{100*(7000 - 4724.2462)}{7000} = 32.5108$

(b) We need to solve the following equation for $t$

$\displaystyle P(t) = 500+6500e^{-0.0431t} = 0.10*7000$

Solving,

$t = 80.7785$

Please check the calculations. Also let me know if you could follow all the steps.

PTDS
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