Dairy Farming

J

john3j

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Nov 18, 2012
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Can anyone please help me figure this out?

"The number of dairy farmers in a particular state who are feeding a new supplement to their milking cows is given by the function W(t)=340(1-e^-0.09t), where t is the number of months and the supplement has been available. How long will it be before 200 farmers are feeding the supplement to their cows?"

If you could provide steps on how to solve this, it would be appreciated.

Thanks,
John
 
What kind of "steps" do you need? The problem is asking you to determine when W(t) is equal to 200. In other words, solve 340(1-e^-0.09t)= 200. While this problem may have come up in a Calculus class, no calculus is required, it is purely an algebra problem.
 
HallsofIvy said:
What kind of "steps" do you need? The problem is asking you to determine when W(t) is equal to 200. In other words, solve 340(1-e^-0.09t)= 200. While this problem may have come up in a Calculus class, no calculus is required, it is purely an algebra problem.
Click to expand...

Thanks HallsofIvy! I ended up with t=0.4505

Appreciate the help!
 
Could you please show exactly how you got that? It does not look at all correct to me.
 
Did you even consider checking your answer? If t= 0.4505, then -.09t= -0.040545. So \(\displaystyle e^(-.09t)= e^(0.040545)= 0.96027. Do you know how to get that? If not, you are missing the whole point of these problems! From that 1- e^(-.09t)= 1- .96027= 0.03973 and so 340(1-e^(-.09t))= 340(.03973)= 13.5, NOT 200.

You entered the problem incorrectly. you want to solve 340(1- e^(-0.09t))= 200 and you entered 340(1- e^(.09)t)= 200. If you do not understand the difference ask your teacher. And stop using internet shortcuts until you have learned how to do such problems yourself! Do you think you are going to be allowed to do that on the exams?\)
 
As stated, you want to solve for t:

\(\displaystyle W(t)=200\)

\(\displaystyle 340\left(1-e^{-0.09t} \right)=200\)

Divide through by 340:

\(\displaystyle 1-e^{-0.09t}=\dfrac{10}{17}\)

Add \(\displaystyle e^{-0.09t}-\dfrac{10}{17}\) to both sides:

\(\displaystyle \dfrac{7}{17}=e^{-0.09t}\)

Invert both sides:

\(\displaystyle \dfrac{17}{7}=e^{0.09t}\)

Convert from exponential to logarithmic form:

\(\displaystyle 0.09t=\ln\left(\dfrac{17}{7} \right)\)

Multiply by \(\displaystyle \dfrac{100}{9}\):

\(\displaystyle t=\dfrac{100}{9}\ln\left(\dfrac{17}{7} \right)\)

Now is the time for a calculator/computer (if you desire a decimal approximation):

\(\displaystyle t\approx9.86\)
 
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