Assuming $$ P \geq 0 $$, suppose that a population develops according to the logistic equation $$ \frac{d P}{d t}=0.01 P-5 \times 10^{-5} P^{2} $$ where $$ t $$ is measured in weeks. Answer the following questions. 1. What is the carrying capacity? Carrying Capacity: 2. What is the value of $$ k $$ ? Answer: $$ k= $$ Ans 3. For what values of $$ P $$ is the population increasing? Answer (in interval notation): 4. For what values of $$ P $$ is the population decreasing? Answer (in interval notation):

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**Step 1. Write the logistic equation in a standard form.** The given logistic equation is $$ \frac{dP}{dt}=0.01P-5 \times 10^{-5}P^{2}. $$ We can factor this as $$ \frac{dP}{dt}=P\left(0.01-5 \times 10^{-5}P\right). $$ --- **Step 2. Find the carrying capacity.** In a logistic model, the carrying capacity $$A$$ is the value of $$P$$ for which the growth term $$0.01-5 \times 10^{-5}P$$ is zero: $$ 0.01-5 \times 10^{-5}P=0. $$ Solve for $$P$$: $$ 5 \times 10^{-5}P=0.01 \quad\Longrightarrow\quad P=\frac{0.01}{5 \times 10^{-5}}. $$ Calculating, we have $$ P=\frac{0.01}{0.00005}=200. $$ Thus, the carrying capacity is $$200$$. --- **Step 3. Determine the value of $$ k $$.** The standard logistic equation is written as $$ \frac{dP}{dt}=kP\left(1-\frac{P}{A}\right). $$ Comparing with $$ \frac{dP}{dt}=0.01P-5 \times 10^{-5}P^{2}=P\left(0.01-5 \times 10^{-5}P\right), $$ we identify: $$ k = 0.01 \quad \text{and} \quad \frac{k}{A}=5 \times 10^{-5}. $$ Since we already found $$A=200$$, we can verify: $$ \frac{0.01}{200}=5 \times 10^{-5}. $$ So, the value of $$ k $$ is $$0.01$$. --- **Step 4. Find the interval for which the population is increasing.** The population is increasing when $$\frac{dP}{dt}>0$$, that is when $$ P\left(0.01-5 \times 10^{-5}P\right)>0. $$ Since $$P \geq 0$$, we consider two factors: - $$P>0$$, and - $$0.01-5 \times 10^{-5}P>0$$. Solve the inequality for the second factor: $$ 0.01 > 5 \times 10^{-5}P \quad\Longrightarrow\quad P < \frac{0.01}{5 \times 10^{-5}} = 200. $$ Thus, the population is increasing for $$0200$$ the expression $$0.01-5 \times 10^{-5}P$$ becomes negative. Hence, the population is decreasing when: $$ P>200. $$ In interval notation, the population is decreasing on: $$ (200,\infty). $$ --- **Final Answers:** 1. Carrying Capacity: $$200$$ 2. $$ k=0.01 $$ 3. The population is increasing on: $$(0,200)$$ 4. The population is decreasing on: $$(200,\infty)$$ **Carrying Capacity:** 200 **Value of $$ k $$:** 0.01 **Population Increasing:** $$0 < P < 200$$ **Population Decreasing:** $$P > 200$$

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