(a) The marginal cost $C'(x)$ and marginal revenue $R'(x)$ are given by $C'(x) = 50 + \dfrac{x}{50}$ and $R'(x) = 60$. The fixed cost is $ ext{₹ } 200$. Determine the maximum profit.

OR

(b) Using the following data, construct Fisher's Ideal Index and show how it satisfies Fact

Question

(a) The marginal cost $C'(x)$ and marginal revenue $R'(x)$ are given by $C'(x) = 50 + \dfrac{x}{50}$ and $R'(x) = 60$. The fixed cost is $	ext{₹ } 200$. Determine the maximum profit.

OR

(b) Using the following data, construct Fisher's Ideal Index and show how it satisfies Factor Reversal Test and Time Reversal Test.

| Commodity | Price in Rupees per unit — Base Year | Price in Rupees per unit — Current Year | Number of units — Base Year | Number of units — Current Year |
| --- | --- | --- | --- | --- |
| A | 6 | 10 | 50 | 56 |
| B | 2 | 2 | 100 | 120 |
| C | 4 | 6 | 60 | 60 |
| D | 10 | 12 | 50 | 24 |
| E | 8 | 12 | 40 | 36 |

Accepted Answer

>

> [!TLDR]
> (a) Set marginal profit to zero: $x=500$, maximum profit $	ext{₹}2300$. (b) Fisher's index $\approx 138.52$, and it passes both the Time Reversal and Factor Reversal tests.
>
> ## Part (a) — Maximum profit
>
> Marginal cost $C'(x) = 50 + \dfrac{x}{50}$, marginal revenue $R'(x) = 60$, fixed cost $	ext{₹}200$.
>
> **Step 1 — marginal profit $= 0$.** $P'(x) = R'(x) - C'(x) = 60 - 50 - \dfrac{x}{50} = 10 - \dfrac{x}{50}.$ Setting $P'(x) = 0$: $x = 500.$ Since $P''(x) = -	frac{1}{50} < 0$, this is a maximum.
>
> **Step 2 — revenue and cost functions.**
> $$R(x) = \int 60\,dx = 60x \quad (R(0)=0).$$
> $$C(x) = \int\left(50 + \frac{x}{50}ight)dx + 200 = 50x + \frac{x^2}{100} + 200.$$
>
> **Step 3 — profit and its maximum value.**
> $$P(x) = R(x) - C(x) = 60x - 50x - \frac{x^2}{100} - 200 = 10x - \frac{x^2}{100} - 200.$$
> At $x = 500$:
> $$P(500) = 10(500) - \frac{500^2}{100} - 200 = 5000 - 2500 - 200 = 2300.$$
>
> **Maximum profit $= 	ext{₹}2300$.**
>
> ## Part (b) — Fisher's Ideal Index and reversal tests
>
> | Commodity | $p_0$ | $p_1$ | $q_0$ | $q_1$ | $p_1q_0$ | $p_0q_0$ | $p_1q_1$ | $p_0q_1$ |
> | --- | --- | --- | --- | --- | --- | --- | --- | --- |
> | A | 6 | 10 | 50 | 56 | 500 | 300 | 560 | 336 |
> | B | 2 | 2 | 100 | 120 | 200 | 200 | 240 | 240 |
> | C | 4 | 6 | 60 | 60 | 360 | 240 | 360 | 240 |
> | D | 10 | 12 | 50 | 24 | 600 | 500 | 288 | 240 |
> | E | 8 | 12 | 40 | 36 | 480 | 320 | 432 | 288 |
> | **Total** | | | | | **2140** | **1560** | **1880** | **1344** |
>
> **Step 1 — Fisher's Ideal Index.**
> $$P_{01}^{F} = \sqrt{\frac{\Sigma p_1 q_0}{\Sigma p_0 q_0}	imes\frac{\Sigma p_1 q_1}{\Sigma p_0 q_1}}	imes 100 = \sqrt{\frac{2140}{1560}	imes\frac{1880}{1344}}	imes 100.$$
> $$= \sqrt{1.3718 	imes 1.3988}	imes 100 = \sqrt{1.9189}	imes 100 = 1.3852	imes 100 = 138.52.$$
>
> **Step 2 — Time Reversal Test** requires $P_{01}	imes P_{10} = 1$ (indices as ratios, without $	imes 100$): …

Commodity	$p_0$	$p_1$	$q_0$	$q_1$	$p_1q_0$	$p_0q_0$	$p_1q_1$	$p_0q_1$
A	6	10	50	56	500	300	560	336
B	2	2	100	120	200	200	240	240
C	4	6	60	60	360	240	360	240
D	10	12	50	24	600	500	288	240
E	8	12	40	36	480	320	432	288
Total					2140	1560	1880	1344

Part (a) — Maximum profit

Part (b) — Fisher's Ideal Index and reversal tests

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