(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:

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