1. (a) A given product can be manufactured at a total cost function \( \mathrm{C}(\mathrm{r})=\mathrm{Re}\left[\frac{x^{2}}{100}+100 \mathrm{x}+40\right] \), where \( x \) is the number of units produced. The price at which each unit can be sold is given by \( \mathrm{P}=\mathrm{Rs} .\left[200-\frac{x}{400}\right] \) (i) Find the total revenue function and maximum revenue. (i) Find the level of output \( x \) for which profit is maximized (iii) Determine the price per unit and the total profit at this level of output (iv) Find the output level \( x \) at which average cost is minimized. (v) Also, show that the marginal cost and the average cost are equal at the level minirining aerage cost

To tackle this problem, we first need to derive the total revenue function and then find the maximum revenue and other related quantities step by step. ### (i) Finding Total Revenue Function \( R(x) \): The total revenue function \( R(x) \) is defined as the product of the price per unit and the number of units sold \( x \). Given the price function: \[ P = 200 - \frac{x}{400} \] The total revenue is: \[ R(x) = P \cdot x = \left(200 - \frac{x}{400}\right) x = 200x - \frac{x^2}{400} \] To find the maximum revenue, we can differentiate \( R(x) \) and set it to zero: \[ R'(x) = 200 - \frac{2x}{400} = 200 - \frac{x}{200} \] Setting the derivative to zero: \[ 200 - \frac{x}{200} = 0 \implies x = 40000 \quad(\text{units}) \] Now, substituting \( x = 40000 \) back into the revenue function to find maximum revenue: \[ R(40000) = 200(40000) - \frac{(40000)^2}{400} = 8000000 - 4000000 = 4000000 \quad(\text{Rs}) \] ### (ii) Finding Level of Output for Maximum Profit: To find the profit function \( \pi(x) \), we subtract the total cost from total revenue: \[ \pi(x) = R(x) - C(x) = \left(200x - \frac{x^2}{400}\right) - \left[\frac{x^2}{100} + 100x + 40\right] \] Simplifying the profit function: \[ \pi(x) = 200x - \frac{x^2}{400} - \frac{x^2}{100} - 100x - 40 \] Combining like terms: \[ \pi(x) = 100x - \left(\frac{4x^2}{400} + \frac{x^2}{400}\right) - 40 = 100x - \frac{5x^2}{400} - 40 = 100x - \frac{x^2}{80} - 40 \] To maximize profit, differentiate and set to zero: \[ \pi'(x) = 100 - \frac{x}{40} \] Setting it to zero: \[ 100 - \frac{x}{40} = 0 \implies x = 4000 \quad(\text{units}) \] ### (iii) Determine Price per Unit and Total Profit at \( x = 4000 \): Substituting \( x = 4000 \) into the price function: \[ P(4000) = 200 - \frac{4000}{400} = 200 - 10 = 190 \quad(\text{Rs}) \] Now, substituting \( x = 4000 \) into the profit function to find total profit: \[ \pi(4000) = 100(4000) - \frac{(4000)^2}{80} - 40 = 400000 - \frac{16000000}{80} - 40 = 400000 - 200000 - 40 = 199960 \quad(\text{Rs}) \] ### (iv) Finding Output Level \( x \) at which Average Cost is Minimized: First, compute the average cost function \( AC(x) \): \[ C(x) = \frac{x^2}{100} + 100x + 40 \Rightarrow AC(x) = \frac{C(x)}{x} = \frac{\frac{x^2}{100} + 100x + 40}{x} = \frac{x}{100} + 100 + \frac{40}{x} \] To find the minimum, differentiate \( AC(x) \): \[ AC'(x) = \frac{1}{100} - \frac{40}{x^2} \] Setting it to zero for minimization: \[ \frac{1}{100} - \frac{40}{x^2} = 0 \implies \frac{40}{x^2} = \frac{1}{100} \implies x^2 = 4000 \implies x = 63.2455 \quad(\text{units}) \] ### (v) Showing Marginal Cost equals Average Cost at Minimum Average Cost: The marginal cost is given by the derivative of the total cost function: \[ MC = C'(x