(a) The marginal cost $C'(x)$ and marginal revenue $R'(x)$ are given by $C'(x)=50+\frac{x}{50}$ and $R'(x)=60$. The fixed cost is ₹ 200. Determine the maximum profit. OR (b) Using Vogel's approximation method, obtain the initial feasible solution of the following transportation problem. | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | | --- | --- | --- | --- | --- | --- | | $O_1$ | 2 | 3 | 11 | 7 | 6 | | $O_2$ | 1 | 0 | 6 | 1 | 1 | | $O_3$ | 5 | 8 | 15 | 9 | 10 | | Demand | 7 | 5 | 3 | 2 | |

(a) The marginal cost C'(x) and marginal revenue R'(x) are given by C'(x)=50+(x)/(50) and R'(x)=60. The fixed

Question 31 of 42

The marginal cost $C'(x)$ and marginal revenue $R'(x)$ are given by $C'(x)=50+\frac{x}{50}$ and $R'(x)=60$ . The fixed cost is ₹ 200. Determine the maximum profit. OR
Using Vogel's approximation method, obtain the initial feasible solution of the following transportation problem.

	$D_1$	$D_2$	$D_3$	$D_4$	Supply
$O_1$	2	3	11	7	6
$O_2$	1	0	6	1	1
$O_3$	5	8	15	9	10
Demand	7	5	3	2

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2024Subjective· 5mImportance★★★★★

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(a) Set $MR=MC\Rightarrow x=500$ ; profit $P(x)=10x-\tfrac{x^2}{100}-200$ gives ₹2300. (b) VAM initial solution costs ₹102.

Part (a) — maximum profit from marginals. Profit is greatest when marginal revenue equals marginal cost.

$R'(x)=C'(x):\quad 60=50+\frac{x}{50}\ \Rightarrow\ \frac{x}{50}=10\ \Rightarrow\ x=500.$

Total revenue $R(x)=\displaystyle\int R'(x)\,dx=\int 60\,dx=60x$ (with $R(0)=0$ ).

Total cost $C(x)=\displaystyle\int C'(x)\,dx=\int\!\left(50+\frac{x}{50}\right)dx=50x+\frac{x^{2}}{100}+k$ ; fixed cost $k=200$ , so $C(x)=50x+\dfrac{x^{2}}{100}+200$ .

Profit:

$P(x)=R(x)-C(x)=60x-\left(50x+\frac{x^{2}}{100}+200\right)=10x-\frac{x^{2}}{100}-200.$

At $x=500$ :

$P(500)=10(500)-\frac{500^{2}}{100}-200=5000-2500-200=2300.$

Part (b) — Vogel's Approximation Method. Balanced problem (supply $=6+1+10=17=$ demand $7+5+3+2$ ).

Working through the penalties (difference of the two smallest costs in each row/column) and allocating to the least-cost cell of the row/column with the largest penalty:

Largest penalty column $D_4$ (=6) $\to$ least cost $O_2$ (=1): allocate $1$ to $(O_2,D_4)$ ; $O_2$ exhausted.
Largest penalty $D_2$ (=5) $\to$ least cost $O_1$ (=3): allocate $5$ to $(O_1,D_2)$ ; $D_2$ met.
Largest penalty $O_1$ (=5) $\to$ least cost $D_1$ (=2): allocate $1$ to $(O_1,D_1)$ ; $O_1$ exhausted. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set MARCH1 markMCQ
$p(x)$
(a) $MR = 0$
(b) $MC - MR = 0$
(c) $MC + MR = 0$
(d) $MC = 0$
›Reveal solutionSolution

Profit is maximum where marginal revenue equals marginal cost: $MR=MC$ , i.e. $MC-MR=0$ .

Profit is $p(x)=R(x)-C(x)$ , where $R$ is total revenue and $C$ is total cost. For a maximum, the first derivative must vanish:

$p'(x) = R'(x) - C'(x) = MR - MC = 0 \;\Rightarrow\; MR = MC.$

…

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CBSE 2025Set MARCH1 markMCQ
$Q. If the marginal revenue of a firm is a constant, then the demand function is :$
(a) $C(x)$
(b) $MR$
(c) $AC$
(d) $MC$
›Reveal solutionSolution

A constant marginal revenue integrates to $R=kx$ (no constant, since $R(0)=0$ ); the demand price $p=R/x=k$ equals $MR$ , option (b).

Integrate the constant $MR$ . Let $MR=k$ . Then

$R=\int MR\,dx=\int k\,dx=kx+C.$

Fix the constant. Revenue is zero when $x=0$ , so $C=0$ and $R=kx$ .

…

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CBSE 2023Set MARCH1 markMCQ
$p(x)$
(a) $MR=0$
(b) $MC-MR=0$
(c) $MC+MR=0$
(d) $MC=0$
›Reveal solutionSolution

Profit is maximised where its derivative is zero: $MR=MC$ , i.e. $MC-MR=0$ .

Profit is defined as revenue minus cost:

$P(x)=R(x)-C(x).$

For a maximum, set the first derivative to zero:

$P'(x)=R'(x)-C'(x)=0\;\Rightarrow\;MR-MC=0\;\Rightarrow\;MC-MR=0,$ …

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CBSE 2023Set MARCH1 markMCQ
$MR=100-9x^2$
(a) $100x-9x^2$
(b) $100-3x^2$
(c) $100+9x^2$
(d) $100x-3x^2$
›Reveal solutionSolution

Total revenue is the integral of marginal revenue; dividing $R$ by $x$ gives the demand (price) function $p=100-3x^2$ .

Given the marginal revenue $MR=100-9x^2$ . Total revenue is:

$R=\int MR\,dx=\int (100-9x^2)\,dx=100x-3x^3+c.$

At $x=0$ revenue is $0$ , so $c=0$ :

$R=100x-3x^3.$

…

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CBSE 2022Set MARCH1 markMCQ
$p(x)$
(a) $MR = 0$
(b) $MC - MR = 0$
(c) $MC + MR = 0$
(d) $MC = 0$
›Reveal solutionSolution

Profit peaks when $MC = MR$ , i.e. $MC - MR = 0$ .

Profit is total revenue minus total cost: $p(x) = R(x) - C(x)$ . For a maximum, differentiate and set the derivative to zero:

$\frac{dp}{dx} = \frac{dR}{dx} - \frac{dC}{dx} = MR - MC = 0.$

…

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CBSE 2022Set MARCH1 markMCQ
$\int \frac{dp}{p} = k \int \frac{dx}{x}$
(a) $-\frac{1}{\eta_d}$
(b) $\eta_d$
(c) $\frac{1}{\eta_d}$
(d) $-\eta_d$
›Reveal solutionSolution

The constant $k = -\dfrac{1}{\eta_d}$ .

The elasticity of demand is defined as

$\eta_d = -\frac{p}{x}\cdot\frac{dx}{dp}.$

Integrate the given relation $\displaystyle\int \frac{dp}{p} = k\int \frac{dx}{x}$ :

$\ln p = k\ln x + \text{constant} \;\Rightarrow\; p = C\,x^{k}.$

Differentiate: $\dfrac{dp}{dx} = Ck\,x^{k-1} = k\cdot\dfrac{p}{x}$ , so $\dfrac{dx}{dp} = \dfrac{x}{k\,p}$ .

…

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CBSE 2020Set MARCH1 markMCQ
$MC = 100\sqrt{x}$
(a) $\dfrac{200}{3x^{1/2}}$
(b) $\dfrac{200}{3}x^{1/2}$
(c) $\dfrac{200}{3}x^{3/2}$
(d) $\dfrac{200}{3x^{3/2}}$
›Reveal solutionSolution

Integrate $MC$ to get $TC$ , fix the constant using $TC = 0$ at $x = 0$ , then divide by $x$ to obtain the average cost $AC = \frac{200}{3}x^{1/2}$ .

Step 1 — Total cost from marginal cost.

$TC = \int MC\,dx = \int 100\sqrt{x}\,dx = 100\int x^{1/2}\,dx = 100\cdot \frac{x^{3/2}}{3/2} = \frac{200}{3}x^{3/2} + c.$

…

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