Q.(a) Solve the following LPP
Maximize $Z=2x_1+5x_2$ subject to the conditions $x_1+4x_2\leq 24$, $3x_1+x_2\leq 21$, $x_1+x_2\leq 9$ and $x_1, x_2\geq 0$.

(a) Corner points $(0,0),(7,0),(6,3),(4,5),(0,6)$; $Z$ values $0,14,27,33,30$; max $Z=33$ at $(4,5)$. (b) General term $\binom{2n}{r}x^{2n-2r}$; $r=n$ gives $\binom{2n}{n}$, which simplifies to the required product.

Part (a) — Graphical solution of the LPP

Maximise $Z=2x_1+5x_2$ subject to $x_1+4x_2\le24$, $3x_1+x_2\le21$, $x_1+x_2\le9$, $x_1,x_2\ge0$.

Step 1 — find the corner points of the feasible region.

• $(0,0)$ — origin.
• $(7,0)$ — from $3x_1+x_2=21$ on the $x_1$-axis (most binding: $7<9$).
• $(0,6)$ — from $x_1+4x_2=24$ on the $x_2$-axis (most binding: $6<9$).
• Intersection of $x_1+4x_2=24$ and $x_1+x_2=9$: subtracting, $3x_2=15\Rightarrow x_2=5,\ x_1=4$, i.e. $(4,5)$ (feasible: $3(4)+5=17\le21$).
• Intersection of $x_1+x_2=9$ and $3x_1+x_2=21$: subtracting, $2x_1=12\Rightarrow x_1=6,\ x_2=3$, i.e. $(6,3)$ (feasible: $6+12=18\le24$).

Step 2 — evaluate $Z=2x_1+5x_2$ at each corner.

Step 3 — choose the maximum. The largest value is $Z=33$ at $(4,5)$.

Part (b) — Term independent of $x$

Given: expansion of $\left(x+\dfrac1x\right)^{2n}$.

Step 1 — general term.

$T_{r+1}=\binom{2n}{r}x^{2n-r}\left(\frac1x\right)^{r}=\binom{2n}{r}x^{2n-2r}.$

Step 2 — condition for independence of $x$.

$2n-2r=0\;\Rightarrow\;r=n.$

Step 3 — the required term. …