Differential Equations (Forestry)

[knowLittle]

Homework Statement

The value of a tract of timber is

$V(t)=100,000 e^{0.8\sqrt{t} }$

where t is the time in years, with t=0 corresponding to 1998.
If money earns interests continuously at 10%, the present value of the timber at any time t is $A(t)= V(t)e^{-.10 t}$
Find the year in which the timber should be harvested to maximize the present value function.

The Attempt at a Solution

When V(0)= 100,000 is the value of timber in 1998.

$A(t)= 100,000 e^{0.8\sqrt{t} }\ e^{-.10 t}$ I assume that to maximize this function I need to make the exponential positive rather than negative.
But, I don't know hot to proceed from here.

 

[SteamKing]

For what values of t is the exponential negative?

If the function to be maximized did not have an exponential term, how would you go about maximizing it?

 

[knowLittle]

For what values of t is the exponential negative?

I wanted to find a value of t such that by addition with 0.8 SQRT(t) would give me a positive exponent.

If the function to be maximized did not have an exponential term, how would you go about maximizing it?

If the function did not have an exponential term, I would try to find a value of T that multiplied by the constant 100,000 would be the greatest value for A.

 

[LCKurtz]

@knowLittle Do you know how to find the maximum/minimum of a function using calculus?

 

[knowLittle]

I know that I can get the relative extrema through differentiation, but I am having lots of problems to get the derivative of this A(t) function.

$\frac{d}{dt} (100,000e^{0.8 \sqrt{t} -.10 t}) =0$

I get :
$0.01t^{1/2} + 0.32 \frac{\sqrt{t}}{t}- 0.12 =0$

and I don't know how to solve for t.

 

[LCKurtz]

I know that I can get the relative extrema through differentiation, but I am having lots of problems to get the derivative of this A(t) function.

$\frac{d}{dt} (100,000e^{0.8 \sqrt{t} -.10 t}) =0$

I get :
$0.01t^{1/2} + 0.32 \frac{\sqrt{t}}{t}- 0.12 =0$

and I don't know how to solve for t.

I don't get that. Please show your steps taking the derivative.

 

[knowLittle]

0= 100,000 u e^u du. 100,000 cancels and u*du is multiplied.

$u={0.8 \sqrt{t} -.10 t}$ and $du= 0.4 t^{-1/2}- 0.10$
$0= [(0.4* 0.8)- 0.10 *0.8 \sqrt{t} - 0.10(0.4) \sqrt{t}+ (0.10^2) t ] e^u$

$0.12 \sqrt{t}- 0.01 t - 0.32=0$

 

[LCKurtz]

0= 100,000 u e^u du. 100,000 cancels and u*du is multiplied.

$u={0.8 \sqrt{t} -.10 t}$ and $du= 0.4 t^{-1/2}- 0.10$

That is $\frac {du}{dt} =0.4 t^{-1/2}- 0.10$

$0= [(0.4* 0.8)- 0.10 *0.8 \sqrt{t} - 0.10(0.4) \sqrt{t}+ (0.10^2) t ] e^u$

$0.12 \sqrt{t}- 0.01 t - 0.32=0$

I have no idea where you got those last two lines. The derivative of $100000e^u$ is $100000e^u\frac {du}{dt}= 100000e^{0.8 \sqrt{t} -.10 t}(0.4 t^{-1/2}- 0.10)$. That will be 0 only if the last factor is 0.

 

[knowLittle]

You are right.

$e^u 0.4 t^{-1/2} = e^u 0.10$

$t=16$
So, the only critical point is at t=16.

So, can I say that the function A(t) reaches maximum profit at year 1998 +16= 2014

 

[knowLittle]

Is my answer correct?

Thank you for all your help so far. Specially, thanks to Kurtz.

 

[LCKurtz]

You are right.

$e^u 0.4 t^{-1/2} = e^u 0.10$

$t=16$
So, the only critical point is at t=16.

So, can I say that the function A(t) reaches maximum profit at year 1998 +16= 2014

Is my answer correct?

Thank you for all your help so far. Specially, thanks to Kurtz.

$t=16$ is correct. But if this is a problem you are going to hand in you should be sure to check that that value of $t$ gives a maximum of the function and not accidentally a minimum. Critical points can be either one or even a saddle point.

 

[knowLittle]

This problem is for personal leisure. But, yes; I am aware of the possibility in that critical point.
I have a problem in finding other test values, though.

I know that I have intervals <-00, 16> and <16, 00+>
If I assume that t>0, then, I only have <0,16>, <16, 00+>

For A(16) = 100 000 * e^( (0.8 * 4) - (.10* 16) ) = 495303.2424
Also, my second derivative test of A at the critical point is negative. So, the function is concave at t=16. Then, my relative maximum is at t=16 or year 2014.