Let's solve the problem step by step: ### **Given:** \[ \frac{dP}{dt} = 2P - \frac{P^2}{50} \] - \(P\): Deer population - Each license allows a hunter to kill **3 deer/year** - Let \(L\) = number of hunting licenses issued With hunting included, the new model: \[ \frac{dP}{dt} = 2P - \frac{P^2}{50} - 3L \] --- ### **(a) Largest Number of Licenses for Survival** **Survival** means the deer population doesn't go extinct, i.e., there is at least one positive equilibrium (\(P^*>0\)). Set \(\frac{dP}{dt} = 0\): \[ 2P - \frac{P^2}{50} - 3L = 0 \] \[ 2P - \frac{P^2}{50} = 3L \] Multiply both sides by 50: \[ 100P - P^2 = 150L \] \[ P^2 - 100P + 150L = 0 \] This is a quadratic equation in \(P\): \[ P^2 - 100P + 150L = 0 \] The equilibrium(s) are real and positive if the discriminant is non-negative and the roots are positive: \[ \Delta = (-100)^2 - 4 \times 1 \times 150L \geq 0 \] \[ 10000 - 600L \geq 0 \] \[ 600L \leq 10000 \] \[ L \leq \frac{10000}{600} = \frac{500}{30} = \frac{50}{3} \approx 16.67 \] **Final answer (a):** \[ \boxed{16 \text{ licenses (since you can't issue a fraction of a license)}} \] --- ### **(b) What happens if this many licenses are issued?** If **16 licenses** are issued: \[ L = 16 \] Plug into the equilibrium equation: \[ P^2 - 100P + 150 \times 16 = 0 \] \[ P^2 - 100P + 2400 = 0 \] Find roots: \[ P = \frac{100 \pm \sqrt{10000 - 4 \times 2400}}{2} \] \[ P = \frac{100 \pm \sqrt{10000 - 9600}}{2} \] \[ P = \frac{100 \pm \sqrt{400}}{2} \] \[ P = \frac{100 \pm 20}{2} \] \[ P_1 = \frac{120}{2} = 60 \] \[ P_2 = \frac{80}{2} = 40 \] So, with 16 licenses, **two equilibria exist: \(P = 40\) and \(P = 60\)**. The lower equilibrium (40) is unstable, the higher (60) is stable. If the population falls below 40, it will decline to extinction; if it starts above 40, it will stabilize at 60. --- ### **(c) Draw a bifurcation diagram** The bifurcation parameter is \(L\) (number of licenses). - For \(L < 16.67\), two equilibria exist (\(P_+\) stable, \(P_-\) unstable). - For \(L = 16.67\), the two equilibria merge (saddle-node bifurcation). - For \(L > 16.67\), no positive equilibria: deer go extinct. **Bifurcation diagram (qualitative sketch):** - **X-axis:** Number of licenses \(L\) - **Y-axis:** Equilibrium deer population \(P\) - For \(L < 16.67\): Two branches, \(P_+\) (upper, stable), \(P_-\) (lower, unstable). - At \(L = 16.67\): Branches meet at \(P = 50\). - For \(L > 16.67\): No equilibria. **Equilibrium formula:** \[ P = 50 \pm \sqrt{2500 - 150L} \] - Plot the **upper branch** (\(P_+\): stable) - Plot the **lower branch** (\(P_-\): unstable) - The branches meet at \(L = 16.67,\, P = 50\). --- ### **Summary Table** | **Part** | **Answer** | |----------|------------| | (a) | \(\boxed{16}\) licenses (max for survival) | | (b) | **Population stabilizes at 60 if above 40; otherwise goes extinct** | | (c) | See bifurcation diagram and explanation above; at \(L = 16.67\), saddle-node bifurcation at \(P = 50\) | If you need a sketch, I can also provide a plot!

**Final Answers:** **(a)** The largest number of licenses that can be issued for the deer population to survive is **16 licenses**. **(b)** If **16 licenses** are issued, the deer population will stabilize at **60 deer** (the higher equilibrium). If fewer licenses are issued, the population stabilizes at a lower positive number. If more than 16 licenses are issued, the population declines to extinction. **(c)** **Bifurcation Diagram Description:** - The x-axis represents the number of issued licenses \(L\). - The y-axis represents the equilibrium deer population \(P\). - For \(L < 16.67\), **two equilibria** exist: - A **stable** equilibrium at \(P \approx 50 + \sqrt{2500 - 150L}\) (upper branch). - An **unstable** equilibrium at \(P \approx 50 - \sqrt{2500 - 150L}\) (lower branch). - At \(L = 16.67\), the two equilibria **merge at \(P = 50\)** (saddle-node bifurcation). - For \(L > 16.67\), **no positive equilibrium** exists; the deer population goes extinct. --- ### **Bifurcation Diagram Sketch:** Below is a qualitative sketch of the bifurcation diagram (please visualize or draw accordingly): ``` Population (P) | | /\ | / \ | / \ | / \ | / \ |--------------------/----------\---------------- (L = 16.67, P = 50) | \ / | \ / | \ / | \ / | \ / | \ / +---------------------------------------------------→ Number of licenses (L) ``` - The **upper branch** (stable) decreases as \(L\) approaches 16.67 and terminates at the saddle-node point. - The **lower branch** (unstable) also converges at the bifurcation point. --- If you need a more precise plot, I recommend plotting the equilibrium formula: \[ P = 50 \pm \sqrt{2500 - 150L} \] for \(L\) in the range [0, 16.67], with the bifurcation at \(L = 16.67\). --- **Summary:** - Max licenses for survival: **16** - Population behavior at that point: stabilizes at **60 deer** - Bifurcation at \(L \approx 16.67\), with the population disappearing for higher licenses.