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Differential Equations (Forestry)

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    3. 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.
     
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  3. Oct 31, 2012 #2

    SteamKing

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    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?
     
  4. Oct 31, 2012 #3
    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 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.
     
  5. Oct 31, 2012 #4

    LCKurtz

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    @knowLittle Do you know how to find the maximum/minimum of a function using calculus?
     
  6. Oct 31, 2012 #5
    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.
     
  7. Oct 31, 2012 #6

    LCKurtz

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    I don't get that. Please show your steps taking the derivative.
     
  8. Oct 31, 2012 #7
    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 ##
     
    Last edited: Oct 31, 2012
  9. Oct 31, 2012 #8

    LCKurtz

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

    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.
     
  10. Oct 31, 2012 #9
    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
     
  11. Nov 2, 2012 #10
    Is my answer correct?

    Thank you for all your help so far. Specially, thanks to Kurtz.
     
  12. Nov 2, 2012 #11

    LCKurtz

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    ##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.
     
  13. Nov 2, 2012 #12
    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.
     
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