Question

A material decays exponentially at 6.8% per week. How much material will be left after one year, if the initial amount was 40000kg. Including the equation model

Answer 1

A material decays exponentially at 6.8% per week .

initial amount of material is  40000 kg .

To find   the equation model and How much material will be left after one year ?

Solution :  

Now we have the exponential decay function given by :

$\large A=I(1-r)^n$

where

r is the rate of decay in decimal per period

So here it is given that   initial amount of material is  40000 kg .

Therefore   

It is given that decays exponentially at 6.8% per week .  

$r=6.8 \% \ \ per \ \ week =\frac{6.8}{100} \ per \ week =0.068 \ \ per \ week$  

Therefore we have values

and

Therefore $A=I(1-r)^n \implies A=40000(1-0.068)^n \implies A =40000(0.932)^n$

Therefore exponential model for the amount of material that will be left after " n weeks "   is :  

$\large \boxed{A=40000(0.932)^n} \ \ \ .... \ \ \ (n \ \ is \ \ the \ \ number \ \ of \ \ week )$

Next we have to find   How much material will be left after one year ?

So   

$1 \ year = 365 \ days = \frac{365}{7} \ weeks$

Therefore $n = \frac{365}{7} \ weeks$

So putting this value of "n " in the model we get  

$\large A=40000(0.932)^{\frac{365}{7}} \implies A = 1016.9937 \approx 1017$

Therefore after one year the amount of material that will be left is :

$\large \boxed{ 1017 \ \ kg}$

Therefore answer is :

Equation model is :

$\large \boxed{A=40000(0.932)^n} \ \ \ .... \ \ \ (n \ \ is \ \ the \ \ number \ \ of \ \ week )$

Material left after one year is :

$\large \boxed{ 1017 \ \ kg}$