1. **Problem 1:** Given inequalities: $x \geq 2.5$, $y \geq 3.5$, and $y \leq -4x + 11.5$. Find max and min of $P = 3x + 2y$. - Find intersection points of the constraints forming the feasible region. - At vertices: - Intersection of $x=2.5$ and $y=3.5$ is $(2.5,3.5)$. - Intersection of $x=2.5$ and $y=-4x + 11.5$ is $(2.5,-4(2.5)+11.5) = (2.5,1.5)$, but $y\geq3.5$ so this point is invalid. - Intersection of $y=3.5$ and $y=-4x+11.5$: $3.5 = -4x +11.5 \Rightarrow -4x= -8 \Rightarrow x=2$. But $x\geq 2.5$, so no. - So vertices are $(2.5,3.5)$ and where else? Because $y \leq -4x + 11.5$, and $y \geq 3.5$, and $x \geq 2.5$, the feasible region is bounded by $(2.5,3.5)$ and corner where $y=3.5$ and $x$ is large until $y$ no longer satisfies; the only vertex is $(2.5,3.5)$. - Evaluate $P$ at $(2.5,3.5)$: $3(2.5)+2(3.5)=7.5 +7=14.5$. No other vertices satisfy all constraints, so max/min both 14.5. 2. **Problem 2:** $x \geq 5$, $y \geq 8$, $y \leq 3x +5$, find max/min $P = x + y$. - Vertices from: - $(5,8)$ - $(5,3(5)+5) = (5,20)$ - Intersection of $y=8$ and $y=3x+5$: $8=3x+5 \Rightarrow 3x=3 \Rightarrow x=1$, but $x \geq 5$ no. - So vertices: $(5,8)$ and $(5,20)$ - Evaluate $P$: - at $(5,8)$: $5+8=13$ - at $(5,20)$: $5+20=25$ - Max $P=25$ at $(5,20)$, min $P=13$ at $(5,8)$. 3. **Problem 3:** $x \geq 2.5$, $y \geq 10.5$, $y \geq 2x + 7.5$, find max/min $P = 2(x + y)$. - Feasible region bounded by: - Line $y=10.5$, line $y=2x+7.5$, both below or equal. - Intersection point: set $10.5=2x +7.5 \Rightarrow 2x=3 \Rightarrow x=1.5$, but $x \geq 2.5$, so point invalid. - Consider vertices: - $(2.5,10.5)$ since $y \geq 10.5$ - $(2.5, 2(2.5) +7.5) = (2.5, 12.5)$ - Check which is higher: $y$ coordinate max is from $y \geq 12.5$ for $x=2.5$ or $y \geq 10.5$, so higher is $y \geq 12.5$. - Evaluate $P=2(x+y)$: - at $(2.5,10.5)$: $2(2.5+10.5) = 2(13) = 26$ - at $(2.5,12.5)$: $2(2.5 + 12.5) = 2(15) = 30$ - Max $P=30$ at $(2.5,12.5)$, min $P=26$ at $(2.5,10.5)$. 4. **Problem 4:** Music vendor with $855$, CDs at $90$ each, cassettes $45$ each. (a) Inequalities: - $90x + 45y \leq 855$ - $1 < x \leq 7$ - $y \geq 3$ (b) Graph plotted with these inequalities shaded. (c) (i) Profit $= 20x + 8y$ (ii) Vertices of feasible region found by intersections: - Intersection $90x + 45y = 855$ and $x=7$: $90(7)+45y=855 \Rightarrow 630 + 45y=855 \Rightarrow 45y=225 \Rightarrow y=5$ - Intersection $90x + 45y=855$ and $y=3$: $90x + 45(3) = 855 \Rightarrow 90x + 135=855 \Rightarrow 90x=720 \Rightarrow x=8$, but $x \leq 7$, so invalid. - Check $(1,3)$, but $x>1$ so take slightly more. - Evaluate profit at vertices: - $(7,5)$: $20(7) + 8(5)=140 + 40=180$ - $(7,3)$: profit $= 20*7 + 8*3=140 + 24=164$ - $(1,3)$ minimal limit: profit $= 20 + 24=44$ - Max profit $=180$ at $(7,5)$ CDs and cassettes. 5. **Problem 5:** Student has $265$, jerseys $25$ each, shirts $35$ each. (a) Inequalities: - $25x + 35y \leq 265$ - $2 \leq x \leq 5$ - $y > 1$ (b) Graph with shading. (c) (i) Profit $= 5x + 8y$ (ii) Intersection points: - At $x=5$, $25(5)+35y \leq 265 \Rightarrow 125 + 35y \leq 265 \Rightarrow 35y \leq 140 \Rightarrow y \leq 4$ - At $y=1$, $25x + 35(1) \leq 265 \Rightarrow 25x + 35 \leq 265 \Rightarrow 25x \leq 230 \Rightarrow x \leq 9.2$, but max $x=5$ - Vertices: $(2,1)$, $(2,4)$, $(5,1)$, $(5,4)$ - Evaluate profit: - $(2,1)$: $5(2)+8(1)=10+8=18$ - $(2,4)$: $10 + 32=42$ - $(5,1)$: $25 + 8=33$ - $(5,4)$: $25 + 32=57$ - Max profit $=57$ at $(5,4)$ 6. **Problem 6:** Butcher has $9800$, cost per cow $900$, per pig $700$. - Number of pigs $<$ number of cows: $y < x$ - Total animals $\\leq 12$: $x + y \leq 12$ - Budget: $900x + 700y \leq 9800$ (b) Graph of inequalities above. (c) Profit per cow $550$, per pig $325$. - Profit $= 550x + 325y$ - Find vertices from intersection: - Between $y=x-1$ (strict inequality approximated), $x+y=12$, $900x+700y=9800$. - Intersection of $x+y=12$ and $y=x-1$: Substitute $y$: $x + (x-1) = 12 \Rightarrow 2x = 13 \Rightarrow x=6.5, y=5.5$ - Check budget at $(6.5,5.5)$: $900*6.5 + 700*5.5 = 5850 + 3850=9700 \leq 9800$ - Intersection budget and $x+y=12$: From $x+y=12$, $y=12-x$, substitute to budget: $900x + 700(12 - x) =9800 \Rightarrow 900x + 8400 - 700x =9800 \Rightarrow 200x =1400 \Rightarrow x=7$, $y=5$ - Check profit: - At $(7,5)$: $550*7 + 325*5 = 3850 + 1625 = 5475$ - At $(6.5,5.5)$: $550*6.5 + 325*5.5 = 3575 + 1787.5 = 5362.5$ - Max profit $5475$ at $(7,5)$ animals. 7. **Problem 7:** Small and large stores, with $x$ small, $y$ large. (a) Inequalities: - $x + y \leq 35$ - $x \geq 10$ - $y > 2x$ (b) Graph accordingly. (c) Income $= 2500x + 4500y$. - Find vertices by intersections: - $x=10$, $y=2x=20$, so point $(10,20)$ - $x=10$, $x + y =35 \Rightarrow y=25$ - $y=2x$, $x + y=35$, substitute $y=2x$: $ x + 2x =35 \Rightarrow 3x=35 \Rightarrow x=11.67$, $y=23.33$ - Values of income at vertices: - $(10,20)$: $2500*10+4500*20=25000+90000=115000$ - $(10,25)$: $25000 + 112500=137500$ - $(11.67,23.33)$: $2500*11.67 + 4500*23.33 =29175 + 104985=134160$ - Maximum income $=137500$ at $(10,25)$ stores. 8. **Problem 8:** Doctor with capsules X and Y to satisfy chemical units. Constraints (assuming given content per capsule is known, e.g., as per table): - System of inequalities for chemical A, B, C: - $a_x x + a_y y \geq 56$ - $b_x x + b_y y \geq 60$ - $c_x x + c_y y \geq 30$ (Exact coefficients not provided; problem setup only.)