you are the manager of a forest products company with land recently planted (t=0) whith a fast-growing species of a pine. the merchantable volume of timber at instant t>0 is given by v(t) =10t+t^-0.01t°
a. what is the maximum volume and when does it occur
b. what harvest time maximizes mean annual increnent ,[v(t)/t] and what is the associated volume?
C. if per -unit cost of harvesting the timber is birr10, price is birr 11 and the discount rate is r= 0.05 , what is the optimal harvest time, volume at optimal harvest time and present value of NB at t=0?
a. To find the maximum volume and when it occurs, we need to find the maximum of the given function v(t) = 10t + t^(-0.01t°).
To do this, we can take the derivative of v(t) with respect to t and set it equal to zero:
dv(t)/dt = 10 - 0.01t°(-0.01t° - 1) = 0
Simplifying this equation:
10 - 0.01t°(-0.01t° - 1) = 0
10 - 0.01t°(-0.01t° - 1) = 0
10 - 0.01t(-0.01t - 1) = 0
10 - 0.0001t² - 0.01t = 0
0.0001t² + 0.01t - 10 = 0
We can solve this quadratic equation for t, either by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-0.01 ± √(0.01² - 4 * 0.0001 * (-10))) / (2 * 0.0001)
t = (-0.01 ± √(0.0001 + 0.004)) / 0.0002
t ≈ (-0.01 ± √0.0041) / 0.0002
Since we are interested in t > 0, we can ignore the negative value.
t ≈ (-0.01 + √0.0041) / 0.0002
t ≈ (-0.01 + 0.064) / 0.0002
t ≈ 0.054 / 0.0002
t ≈ 270
The maximum volume occurs at t = 270, and the maximum volume is v(270) = 10(270) + (270^(-0.01(270°))) = 2700 + (270^(-2.7)).
b. To find the harvest time that maximizes mean annual increment [v(t)/t], we need to find the maximum of the function [v(t)/t].
We can rewrite the function [v(t)/t] as [10t/t + t^(-0.01t°)/t]:
[v(t)/t] = 10 + t^(-0.01t°-1)
To find the maximum, we need to take the derivative of [v(t)/t] with respect to t and set it equal to zero:
d[v(t)/t]/dt = -0.01t°(-0.01t° - 1) = 0
Simplifying this equation:
-0.01t°(-0.01t° - 1) = 0
-0.01t(-0.01t - 1) = 0
-0.0001t² - 0.01t = 0
t(-0.0001t - 0.01) = 0
This equation has two solutions: t = 0 and t ≈ -100. Since we are interested in t > 0, we can ignore the t = 0 solution.
Therefore, the harvest time that maximizes mean annual increment is t ≈ -100.
To find the associated volume, we can substitute this value into the original volume function:
v(-100) = 10(-100) + (-100)^(-0.01(-100°))
v(-100) = -1000 + (-100)^(-10)
(Note: Since the given volume function is not defined for negative values of t, the result might not be meaningful.)
c. To find the optimal harvest time, we need to consider the present value of net benefits (PVNB) at t=0.
The present value of net benefits is given as:
PVNB = ∫[v(t)*(1+r)^(-t)]dt
Substituting the given volume function into the integral:
PVNB = ∫[(10t + t^(-0.01t°))*(1+0.05)^(-t)]dt
To find the optimal harvest time, we need to maximize this integral. This can be done using numerical methods such as integration techniques or optimization algorithms.
Once the optimal harvest time is found, we can determine the volume at that time by substituting the value of t into the volume function v(t).
Note: Without specific values for t or additional information, it is not possible to provide the exact optimal harvest time, volume at optimal harvest time, or present value of NB at t=0.
To find the answers to the questions, we need to apply some calculations to the given formula. Let's go step by step:
a. To find the maximum volume and when it occurs, we need to find the derivative of the volume function and set it to zero.
v(t) = 10t + t^(-0.01t)
Differentiating v(t) with respect to t:
v'(t) = 10 - 0.01t^(-0.01t-1) * (0.01t * ln(t) + 1)
Setting v'(t) = 0:
10 - 0.01t^(-0.01t-1) * (0.01t * ln(t) + 1) = 0
Now, we can solve this equation numerically to find the value of t when v'(t) = 0.
b. To find the harvest time that maximizes the mean annual increment [v(t)/t] and the associated volume, we need to calculate [v(t)/t] and find its maximum value.
[v(t)/t] = (10t + t^(-0.01t))/t = 10 + t^(-0.01t-1)
Differentiating [v(t)/t] with respect to t:
[v(t)/t]' = 0 - 0.01t^(-0.01t-2) * (0.01t * ln(t) + 1) + (-0.01t-1 * ln(t)) * (10 + t^(-0.01t-1))
Setting [v(t)/t]' = 0:
0 - 0.01t^(-0.01t-2) * (0.01t * ln(t) + 1) + (-0.01t-1 * ln(t)) * (10 + t^(-0.01t-1)) = 0
Now, we can solve this equation numerically to find the value of t that maximizes [v(t)/t].
c. To find the optimal harvest time, volume at the optimal harvest time, and the present value of Net Benefits (NB) at t=0, we need to calculate the Net Benefits (NB) formula.
NB = (price * v(t) - cost) / (1 + r)^t
where price is birr 11, cost is birr 10, and r (discount rate) is 0.05.
By substituting the values into the formula, we can calculate the Net Benefits for each harvest time and find the time that maximizes the Net Benefits. Then, we can calculate the associated volume by plugging the optimal harvest time into the volume function v(t).
Please note that due to the complexity of the equations and the need for numerical calculations, it would be more suitable to use mathematical software or tools to find the exact values.