Q.Refer to Exercise 7 above. Find the maximum value of $Z$.

Plot the feasible region from the four constraints, find its corner points, then evaluate $Z=x+2y$ at each corner — the maximum/minimum of a linear objective over a bounded polygon always occurs at a corner.

Constraints: $x+2y\ge100$, $2x-y\le0$ (i.e. $y\ge2x$), $2x+y\le200$, $x\ge0$, $y\ge0$.

Finding the corner points by intersecting the boundary lines in pairs:

• $x=0$ with $x+2y=100$: gives $(0,50)$.
• $x+2y=100$ with $y=2x$: substituting, $x+4x=100\Rightarrow x=20,\,y=40$: gives $(20,40)$.
• $y=2x$ with $2x+y=200$: substituting, $2x+2x=200\Rightarrow x=50,\,y=100$: gives $(50,100)$.
• $2x+y=200$ with $x=0$: gives $(0,200)$.

(One can check each of these four points satisfies all the constraints simultaneously, and that they are the actual vertices of the bounded feasible region — a quadrilateral.)

Evaluating $Z=x+2y$ at each corner:

This is a linear programming problem: set up the demand and investment constraints, plot the feasible region, and evaluate the profit at each corner point — the maximum occurs at $(200,50)$ giving a profit of Rs $11{,}50{,}000$.

Let $x$ = number of desktop models, $y$ = number of portable models to stock.

Objective (maximize profit): $Z=4500x+5000y$

Constraints:

• Demand: $x+y\le250$
• Investment: $25000x+40000y\le70{,}00{,}000$, which simplifies (dividing by $5000$) to $5x+8y\le1400$
• Non-negativity: $x\ge0,\ y\ge0$

Corner points of the feasible region:

• $(0,0)$
• $x$-intercept of demand line: $(250,0)$ — check investment: $5(250)=1250\le1400$ ✓, so this is a genuine corner.
• $y$-intercept of investment line: $8y=1400\Rightarrow y=175$, giving $(0,175)$ — check demand: $175\le250$ ✓ (this is the binding constraint here, since the demand line would allow $y=250$ but investment caps it at $175$).
• Intersection of $x+y=250$ and $5x+8y=1400$: from the first, $x=250-y$; substitute: $5(250-y)+8y=1400\Rightarrow1250+3y=1400\Rightarrow y=50,\ x=200$. Point: $(200,50)$. …