1. **Problem 1: Determine the feasible solution of the system of linear inequalities:** Given inequalities: $$2x + y \geq 0$$ $$x - 2y > 0$$ $$x \geq 0, \quad y \geq 0$$ 2. **Understanding the inequalities:** - The first inequality means the point $(x,y)$ lies on or above the line $2x + y = 0$. - The second inequality means the point lies strictly above the line $x - 2y = 0$. - The last two inequalities restrict the solution to the first quadrant (including axes). 3. **Rewrite inequalities for graphing:** - From $2x + y \geq 0$, we get $y \geq -2x$. - From $x - 2y > 0$, we get $x > 2y$ or equivalently $y < \frac{x}{2}$. 4. **Feasible region:** - Since $x \geq 0$ and $y \geq 0$, we are in the first quadrant. - The inequality $y \geq -2x$ is always true in the first quadrant because $y \geq 0$ and $-2x \leq 0$. - The inequality $y < \frac{x}{2}$ restricts $y$ to be below the line $y = \frac{x}{2}$. 5. **Conclusion:** The feasible region is all points in the first quadrant where $y$ is less than $\frac{x}{2}$. --- 6. **Problem 2(a): Formulate DMC’s problem as a linear program.** Let: - $x$ = number of four-color presses produced - $y$ = number of two-color presses produced Constraints: - Roller constraint: $16x + 8y \leq 100$ - Gear cutting: $30x + 12y \leq 160$ - Polishing: $8x + 3y \leq 40$ - Minimum production: $x \geq 2$, $y \geq 2$ Objective function (maximize profit): $$\max Z = 24000x + 10000y$$ --- 7. **Problem 2(b): Solve graphically and explain optimal solution.** - Plot the constraints on the $xy$-plane. - Identify the feasible region satisfying all constraints. - Evaluate the objective function $Z$ at each vertex of the feasible region. Vertices are found by solving intersections of constraints: - Intersection of roller and gear cutting: $$16x + 8y = 100$$ $$30x + 12y = 160$$ Multiply first by 3 and second by 2 to align $y$ coefficients: $$48x + 24y = 300$$ $$60x + 24y = 320$$ Subtract: $$\cancel{48x} + 24y - \cancel{48x} - 24y = 300 - 320$$ $$12x = -20 \Rightarrow x = -\frac{20}{12} = -\frac{5}{3}$$ Negative $x$ is invalid due to $x \geq 2$, so no feasible intersection here. - Check intersection of roller and polishing: $$16x + 8y = 100$$ $$8x + 3y = 40$$ Multiply second by 8 and first by 3: $$48x + 24y = 300$$ $$64x + 24y = 320$$ Subtract: $$\cancel{48x} + 24y - \cancel{48x} - 24y = 300 - 320$$ $$16x = -20 \Rightarrow x = -\frac{20}{16} = -1.25$$ Again invalid. - Check intersection of gear cutting and polishing: $$30x + 12y = 160$$ $$8x + 3y = 40$$ Multiply second by 4: $$30x + 12y = 160$$ $$32x + 12y = 160$$ Subtract: $$\cancel{30x} + 12y - \cancel{30x} - 12y = 160 - 160$$ $$2x = 0 \Rightarrow x = 0$$ But $x \geq 2$ so no feasible intersection here. - Check corners at minimum production: At $x=2, y=2$: Rollers: $16(2) + 8(2) = 32 + 16 = 48 \leq 100$ (OK) Gear cutting: $30(2) + 12(2) = 60 + 24 = 84 \leq 160$ (OK) Polishing: $8(2) + 3(2) = 16 + 6 = 22 \leq 40$ (OK) - Check if increasing $x$ or $y$ is possible within constraints. - Maximize profit by increasing $x$ and $y$ within constraints. - The limiting constraint is rollers: $16x + 8y \leq 100$. - For $y=2$, max $x$ is: $$16x + 8(2) \leq 100 \Rightarrow 16x + 16 \leq 100 \Rightarrow 16x \leq 84 \Rightarrow x \leq 5.25$$ - For $x=2$, max $y$ is: $$16(2) + 8y \leq 100 \Rightarrow 32 + 8y \leq 100 \Rightarrow 8y \leq 68 \Rightarrow y \leq 8.5$$ - Check gear cutting for $x=5.25, y=2$: $$30(5.25) + 12(2) = 157.5 + 24 = 181.5 > 160$$ Not feasible. - Check gear cutting for $x=2, y=8.5$: $$30(2) + 12(8.5) = 60 + 102 = 162 > 160$$ Not feasible. - Gear cutting is more restrictive. - For $y=2$, max $x$ from gear cutting: $$30x + 12(2) \leq 160 \Rightarrow 30x + 24 \leq 160 \Rightarrow 30x \leq 136 \Rightarrow x \leq 4.53$$ - For $x=2$, max $y$ from gear cutting: $$30(2) + 12y \leq 160 \Rightarrow 60 + 12y \leq 160 \Rightarrow 12y \leq 100 \Rightarrow y \leq 8.33$$ - Check polishing for $x=4.53, y=2$: $$8(4.53) + 3(2) = 36.24 + 6 = 42.24 > 40$$ Not feasible. - For $y=2$, max $x$ from polishing: $$8x + 3(2) \leq 40 \Rightarrow 8x + 6 \leq 40 \Rightarrow 8x \leq 34 \Rightarrow x \leq 4.25$$ - For $x=2$, max $y$ from polishing: $$8(2) + 3y \leq 40 \Rightarrow 16 + 3y \leq 40 \Rightarrow 3y \leq 24 \Rightarrow y \leq 8$$ - So polishing limits $x$ to 4.25 when $y=2$. - Check rollers for $x=4.25, y=2$: $$16(4.25) + 8(2) = 68 + 16 = 84 \leq 100$$ OK. - Check gear cutting for $x=4.25, y=2$: $$30(4.25) + 12(2) = 127.5 + 24 = 151.5 \leq 160$$ OK. - So point $(4.25, 2)$ is feasible. - Check profit at $(4.25, 2)$: $$Z = 24000(4.25) + 10000(2) = 102000 + 20000 = 122000$$ - Check profit at $(2, 8)$ (max $y$ at $x=2$): $$Z = 24000(2) + 10000(8) = 48000 + 80000 = 128000$$ - Check if $(2,8)$ is feasible: Rollers: $16(2) + 8(8) = 32 + 64 = 96 \leq 100$ (OK) Gear cutting: $30(2) + 12(8) = 60 + 96 = 156 \leq 160$ (OK) Polishing: $8(2) + 3(8) = 16 + 24 = 40 \leq 40$ (OK) - So $(2,8)$ is feasible and yields higher profit. - Check profit at $(2,2)$: $$Z = 24000(2) + 10000(2) = 48000 + 20000 = 68000$$ - Check profit at $(4.25, 2)$ is less than at $(2,8)$. - Check profit at $(2,8)$ is highest among tested vertices. **Optimal solution:** Produce 2 four-color presses and 8 two-color presses weekly. **Maximum profit:** 128000. --- **Summary:** - Problem 1 feasible region is the first quadrant below $y = \frac{x}{2}$. - Problem 2 linear program formulated and solved graphically. - Optimal production is $x=2$, $y=8$ with maximum profit 128000.