Question 1170424
Let's break down each part of the problem.

**(a) Teddy J's Dish Washing Liquid**

Given:
* Demand function: $q = 4000 - 250p$
* Total cost function: $C(q) = 500 + 0.2q$

**(i) Total Revenue (R(q))**

* First, solve the demand function for $p$:
    * $250p = 4000 - q$
    * $p = \frac{4000 - q}{250} = 16 - \frac{q}{250}$
* Total revenue is given by $R(q) = pq$:
    * $R(q) = \left(16 - \frac{q}{250}\right)q$
    * $R(q) = 16q - \frac{q^2}{250}$

**(ii) Profit Function (Π(q))**

* Profit is total revenue minus total cost:
    * $\Pi(q) = R(q) - C(q)$
    * $\Pi(q) = \left(16q - \frac{q^2}{250}\right) - (500 + 0.2q)$
    * $\Pi(q) = 16q - \frac{q^2}{250} - 500 - 0.2q$
    * $\Pi(q) = -\frac{q^2}{250} + 15.8q - 500$

**(iii) Profit at q = 500**

* To determine if profit is increasing or decreasing, we need to find the derivative of the profit function and evaluate it at $q = 500$:
    * $\Pi'(q) = -\frac{2q}{250} + 15.8 = -\frac{q}{125} + 15.8$
    * $\Pi'(500) = -\frac{500}{125} + 15.8 = -4 + 15.8 = 11.8$
* Since $\Pi'(500) > 0$, the profit is increasing when Teddy J produces 500 bottles.

**(iv) Profit Maximization**

* To maximize profit, we set the derivative of the profit function to zero:
    * $\Pi'(q) = -\frac{q}{125} + 15.8 = 0$
    * $\frac{q}{125} = 15.8$
    * $q = 15.8 \cdot 125 = 1975$
* Teddy J should produce 1975 bottles to maximize profits.

**(b) Firm's Average Cost Function**

Given:
* Average cost function: $A(q) = \frac{125}{q} + \frac{q}{16} - 4$

**(i) Output for Minimum Average Costs**

* To find the minimum average cost, we take the derivative of $A(q)$ and set it to zero:
    * $A'(q) = -\frac{125}{q^2} + \frac{1}{16} = 0$
    * $\frac{125}{q^2} = \frac{1}{16}$
    * $q^2 = 125 \cdot 16 = 2000$
    * $q = \sqrt{2000} = 20\sqrt{5} \approx 44.72$
* The level of output for minimum average costs is approximately 44.72.

**(ii) Range for Decreasing Average Costs**

* Average costs are decreasing when $A'(q) < 0$:
    * $-\frac{125}{q^2} + \frac{1}{16} < 0$
    * $\frac{1}{16} < \frac{125}{q^2}$
    * $q^2 < 125 \cdot 16 = 2000$
    * $q < \sqrt{2000} = 20\sqrt{5} \approx 44.72$
* Average costs are decreasing when $0 < q < 20\sqrt{5}$.

**(iii) Practically Feasible Decreasing Range**

* Output must be a positive value.
* The practically feasible range for decreasing average costs is $0 < q < 20\sqrt{5}$.

**(iv) Total Cost Function (C(q))**

* Total cost is average cost multiplied by output:
    * $C(q) = q \cdot A(q)$
    * $C(q) = q \left(\frac{125}{q} + \frac{q}{16} - 4\right)$
    * $C(q) = 125 + \frac{q^2}{16} - 4q$

**(v) Output for Minimum Total Costs**

* To find the minimum total cost, we take the derivative of $C(q)$ and set it to zero:
    * $C'(q) = \frac{2q}{16} - 4 = \frac{q}{8} - 4 = 0$
    * $\frac{q}{8} = 4$
    * $q = 32$
* The level of output for minimum total costs is 32.