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

We are given that   Cost:   C(x) = Re[ x²/100 + 100x + 40 ]           (1)   Price:   P(x) = 200 – x/400                (2) Since the cost function’s real part is taken, we may simply work with   C(x) = x²/100 + 100x + 40. Also, note that x represents the number of units produced (and sold). Below we solve each part. ───────────────────────── Part (i): Total Revenue Function and Maximum Revenue The total revenue (R) is given by the product of the number of units and the price per unit:   R(x) = x · P(x) = x (200 – x/400). Thus,   R(x) = 200x – (x²)/400. To find the maximum revenue, differentiate R(x) with respect to x and set equal to zero. Differentiate:   R′(x) = 200 – (2x)/400. Since (2x)/400 = x/200, we have   R′(x) = 200 – x/200. Set R′(x) = 0:   200 – x/200 = 0  ⟹  x/200 = 200  ⟹  x = 200 × 200 = 40,000. Now, substitute x = 40,000 back into R(x) to get the maximum revenue:   R(40,000) = 200(40,000) – (40,000)²/400 Compute:   200 × 40,000 = 8,000,000,   (40,000)² = 1,600,000,000, and dividing by 400 gives 4,000,000. Thus,   R(40,000) = 8,000,000 – 4,000,000 = 4,000,000. So, the total revenue function is   R(x) = 200x – x²/400, and the maximum revenue is Rs. 4,000,000 when x = 40,000 units. ───────────────────────── Part (ii): Level of Output for Maximum Profit Profit, π(x), is given by revenue minus cost:   π(x) = R(x) – C(x). Substitute the expressions:   π(x) = [200x – x²/400] – [x²/100 + 100x + 40]. Let’s combine like terms. 1. x‐terms:   200x – 100x = 100x. 2. x² terms:   – x²/400 – x²/100. Since 1/100 = 4/400, we have   – (x²/400 + 4x²/400) = – 5x²/400 = – x²/80. 3. Constant term is –40. Thus,   π(x) = 100x – (x²)/80 – 40. Differentiate π(x) with respect to x:   π′(x) = 100 – (2x)/80 = 100 – x/40. Set π′(x) equal to zero for maximum profit:   100 – x/40 = 0  ⟹  x/40 = 100  ⟹  x = 100 × 40 = 4,000. So, the profit is maximized when 4,000 units are produced and sold. ───────────────────────── Part (iii): Price per Unit and Total Profit at x = 4,000 The price per unit when x = 4,000 is given by (2):   P(4000) = 200 – (4000)/400 = 200 – 10 = 190. Now, calculate the profit at x = 4,000 using the profit function:   π(4000) = 100(4000) – (4000)²/80 – 40. Compute:   100 × 4000 = 400,000,   (4000)² = 16,000,000, so (16,000,000)/80 = 200,000. Therefore,   π(4000) = 400,000 – 200,000 – 40 = 200,000 – 40 = 199,960. So at 4,000 units, the price per unit is Rs. 190 and the total profit is Rs. 199,960. ───────────────────────── Part (iv): Output Level at Which Average Cost Is Minimized The Average Cost (AC) is defined as:   AC(x) = C(x) / x = [x²/100 + 100x + 40] / x. Simplify by dividing each term by x:   AC(x) = (x/100) + 100 + (40)/x. To minimize AC(x), differentiate with respect to x and set the derivative equal to zero. Differentiate:   d/dx [AC(x)] = (1/100) + 0 – 40/x² = (1/100) – 40/x². Set the derivative equal to zero:   1/100 – 40/x² = 0  ⟹  40/x² = 1/100  ⟹  x² = 40 × 100 = 4000. Thus,   x = √4000 = 20√10   (units). So, the average cost is minimized when x = 20√10 units (approximately 63.25 units). ───────────────────────── Part (v): Equality of Marginal Cost and Average Cost at the Minimum AC First, find the marginal cost (MC), which is the derivative of the total cost with respect to x. We have   C(x) = x²/100 + 100x + 40. Differentiate with respect to x:   MC = C′(x) = (2x)/100 + 100 = x/50 + 100. The average cost is, as before,   AC(x) = (x/100) + 100 + 40/x. Now, we wish to show that when x = 20√10 (the minimizing value for AC), we have MC = AC. First, compute MC at x = 20√10:   MC(20√10) = (20√10)/50 + 100 = (2√10)/5 + 100. Note that (20/50) = 0.4, so you can also write   MC(20√10) = 0.4√10 + 100. Now compute AC at x = 20√10:   AC(20√10) = (20√10)/100 + 100 + 40/(20√10). Simplify:   (20√10)/100 = 0.2√10,   40/(20√10) = 2/√10. Thus,   AC(20√10) = 0.2√10 + 100 + 2/√10. Now, we want to check that   0.4√10 + 100 ? 0.2√10 + 100 + 2/√10. Subtract 100 from both sides; the question is whether   0.4√10 = 0.2√10 + 2/√10. Subtract 0.2√10 from both sides:   0.2√10 = 2/√10. Multiply both sides by √10:   0.2 (√10)² = 2  ⟹  0.2 × 10 = 2  ⟹  2 = 2. This is true, so we have shown that MC = AC at x = 20√10, the level of output which minimizes the average cost. ───────────────────────── Summary of Answers (i) Total Revenue Function and Maximum Revenue:   R(x) = 200x – x²/400.   Maximum revenue of Rs. 4,000,000 occurs when x = 40,000. (ii) Profit is maximized when x = 4,000 units. (iii) At x = 4,000, the unit price is Rs. 190 and the total profit is Rs. 199,960. (iv) The average cost is minimized when x = 20√10 units (approximately 63.25). (v) At x = 20√10 the marginal cost, MC = x/50 + 100, equals the average cost, AC = x/100 + 100 + 40/x. Any answer equivalent to the one detailed above is correct.