Question

can anyone help me with a differential equation problem?

Answer 1

The value of a tract of timber is v(t)=\[100,000e ^{.8\sqrt{t}}\] where t is the time in years, with t=0 corresponding to 1998. If money earns interest continuously at 10%, the present value of the timber at any time t is \[a(t)= v(t)e ^{-.10t}\] find the year in which the timber should be harvested to maximize the present value function.

Answer 2

Have you tried differentiating \(a(t)\) to find its critical points?

Answer 3

how would i take the derivative of that?

Answer 4

Well, we know that we have:
$$
\Large{
v(t) = 100,000 \cdot e^{0.8 \cdot \sqrt{t}} \\
a(t) = v(t) \cdot e^{-0.1 \cdot t} \\
a(t) = 100,000 \cdot e^{0.8 \cdot \sqrt{t}} \cdot e^{-0.1\cdot t} =\\
= 100,000 \cdot e^{0.8 \cdot \sqrt{t} - 0.1 \cdot t}\\
}
\large{
a'(t) = 100,000 \cdot e^{0.8 \cdot \sqrt{t} - 0.1 \cdot t} \cdot (0.8 \cdot \sqrt{t} - 0.1 \cdot t)' = \\
= 100,000 \cdot e^{0.8 \cdot \sqrt{t} - 0.1 \cdot t} \cdot ( \frac{0.8}{2 \cdot \sqrt{t}} - 0.1) = \\
= 100,000 \cdot e^{0.8 \cdot \sqrt{t} - 0.1 \cdot t} \cdot (\frac{0.4}{\sqrt{t}} - 0.1)
}
$$If it is not clear let me know I'll help you through it =)
Can you go on from here?

Answer 5

thank you that's very helpful :) am i just solving for t now?

Answer 6

i tried solving for t but I got stuck :/

Answer 7

Sorry, my internet got disconnected.

Answer 8

Ok, let's see. We want to find the maximum point of the derivative. This point is a 'critical point' as SithsAndGiggles has mentioned. That means it is a point where the function 'stops changing'. (in our case stops raising).
Because those critical points are where the function stops changing, the derivative at those points will be 0. So lets find the critical points by solving:
$$
a'(t) = 0 \\
\Large 100,000 \cdot e^{0.8 \cdot \sqrt{t} - 0.1 \cdot t} \cdot (\frac{0.4}{\sqrt{t}} - 0.1) = 0
$$ Can you solve it?

Answer 9

yes i got 16. thank you very much! :)

Answer 10

Good =)
Is it all clear? if you have any questions don't hesitate to ask =)