这是一份manchester曼切斯特大学ECON60782T作业代写的成功案例

Solve for the dynamically efficient resource allocation for $r=0,0.05$, $0.10,0.2$, and $0.5$. Note that from the exact solution we have $q_{0}=$ $\left(b Q_{\text {tot }}+r(a-c)\right) / b(2+r)$, and $q_{1}=Q_{\text {tot }}-q_{0}$.

Build a table with columns showing ” $r, “{ }^{” } q_{0} “{ }^{} q_{1}, “$ “PV of marginal profit period $0^{” *}$ and ” $P V$ of marginal profit period $1 . “$ There will be five rows, one for each ” $r^{47}$ value above. Confirm that Hotelling”s rule is satisfied in each case. Provide a brief interpretation of the impacts of rising discount rates on the dynamically optimal price and quantity path over time.

Now suppose that $Q_{\text {tot }}=70$. Build a second table like the one in 6 a above. Provide a brief narrative economic interpretation of the impact of a smaller resource stock on the dynamically efficient allocation of the resource, as well as prices and marginal profit.

ECON60782T COURSE NOTES ：

Suppose that $X(t)$ is the stock or biomass of economically valuable fish at time $t$, and $F(X)$ is the biological growth function for the stock over time (from a calculus standpoint, think of this as the instantaneous growth rate of $X$, equal to $d X / d t)$. $F(X)$ reflects the rate of net recruitment (number of new fish entering a fishery, net of fish removed from the fishery). Suppose that for the particular species in question $F(X)$ can be described by a logistic function:
$$
F(X)=r X(l-X / k)
$$
Note that $r$ is interpreted as the rate of growth of $X$ when the stock $X$ is nearly zero. Note that $k$ is interpreted as the maximum value for the stock (carrying capacity) for a given habitat. Thus:
$$
X=k \rightarrow F(X)=0
$$