Exercise:- You are the of a forest products company with land recently planted (1=0 )with a fast-growing species of pure. The merchmoable of timber t>0 at give by V(t)= 10t +t^2-0.01t^3
A.What the maximum volume and at when does it occur?
(b) Wh harvest time miximizes mean annual increment |V(1)/t| and what is to the associated volume?
(C). If permit sest of having the tubiri Biry 10 price is Birr 11 and the dismant rate 05: what is the optional larvest time, selame at optusal barsest time and present value of Nat
A. To find the maximum volume, we need to find the maximum point of the function V(t). We can do this by taking the derivative of V(t) and setting it equal to zero.
V'(t) = 10 + 2t - 0.03t^2
Setting V'(t) = 0, we have:
0 = 10 + 2t - 0.03t^2
Rearranging and factoring, we get:
0 = -0.03t^2 + 2t + 10
Solving this quadratic equation, we find two possible solutions for t: -109.796 and 169.796. Since time cannot be negative, the maximum volume occurs at t = 169.796.
To find the maximum volume, we substitute t = 169.796 into V(t):
V(169.796) = 10(169.796) + (169.796)^2 - 0.01(169.796)^3
V(169.796) = 1697.96 + 28824.08 - 486989.67
V(169.796) = -457467.63
Therefore, the maximum volume is -457467.63 and it occurs at t = 169.796.
B. To find the harvest time that maximizes the mean annual increment |V(1)/t|, we need to maximize the expression V(1)/t. We can do this by taking the derivative of V(1)/t with respect to t and setting it equal to zero.
(V(1)/t)' = [V'(1)t - V(1)]/t^2
Setting (V(1)/t)' = 0, we have:
0 = [V'(1)t - V(1)]/t^2
Multiplying through by t^2, we get:
0 = V'(1)t - V(1)
Solving for t, we have:
t = V(1)/V'(1)
Substituting in the values for V(1) and V'(1), we have:
t = (-125.11)/(10 + 2(1) - 0.03(1)^2)
t = (-125.11)/(10 + 2 - 0.03)
t = (-125.11)/(12.97)
t ≈ -9.63
Since time cannot be negative, the harvest time that maximizes the mean annual increment is t = -9.63.
To find the associated volume, we substitute t = -9.63 into V(t):
V(-9.63) = 10(-9.63) + (-9.63)^2 - 0.01(-9.63)^3
V(-9.63) = -96.3 + 92.4969 + 221.9478
V(-9.63) = 218.1447
Therefore, the associated volume is 218.1447.
C. The optimal harvest time can be determined by finding the time that maximizes the present value of net revenue (PVNR). The present value (PV) at time t is given by:
PV(t) = p(1 + r)^(-t)
where p is the price (Birr 11), r is the discount rate (0.05), and t is the time.
The net revenue (NR) at time t is given by:
NR(t) = V(t) - c
where c is the cost of harvesting.
The present value of net revenue (PVNR) is given by:
PVNR(t) = PV(t) * NR(t)
To find the optimal harvest time, we can numerically calculate the PVNR for different time values and determine the time that results in the highest PVNR.
For example, we can plug in different time values into the equations above and calculate the PVNR. The time that results in the highest PVNR is the optimal harvest time.
The "selame at optusal barsest time" is unclear and needs clarification before it can be answered. The present value of Nat is also unclear and needs additional information or definitions.
A. To find the maximum volume and when it occurs, we need to differentiate the volume equation V(t) = 10t + t^2 - 0.01t^3 with respect to time t, and then find the values of t where this derivative equals zero.
Taking the derivative of V(t), we get:
V'(t) = 10 + 2t - 0.03t^2
Setting V'(t) equal to zero and solving for t:
10 + 2t - 0.03t^2 = 0
This equation is a quadratic equation, so let's solve it. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c:
t = (-(2) ± √((2)^2 - 4(-0.03)(10))) / (2(-0.03))
Simplifying further:
t = (-2 ± √(4 + 1.2)) / (-0.06)
t = (-2 ± √5.2) / (-0.06)
Calculating the values:
t ≈ -8.88 or t ≈ 21.88
Since we are dealing with time, we discard the negative value. Therefore, the maximum volume occurs at t ≈ 21.88.
B. To maximize the mean annual increment |V(1)/t|, we need to find the value of t that maximizes this expression. Let's find the value of t that makes the derivative of |V(1)/t| equal to zero.
Using the quotient rule, the derivative of |V(1)/t| is given by:
(|V(1)|t - V(1)) / t^2
Setting this derivative equal to zero and solving for t:
(|V(1)|t - V(1)) / t^2 = 0
Since we're looking for a maximum, we'll consider only positive values of t. We can rewrite the equation as:
|V(1)|t - V(1) = 0
Simplifying further:
|V(1)|t = V(1)
Since we're maximizing, let's consider two cases:
1. V(1) > 0: In this case, we have t = V(1) / |V(1)| = 1.
2. V(1) < 0: In this case, we have t = V(1) / |V(1)| = -1.
The associated volume in both cases is V(1).
C. It seems the question is cut off at the end. Could you please provide the complete information so that I can help you?