1. **Problem Statement:** We have two brands of oil, X and Y, made from components A, B, and C. Component stores: A=50, B=40, C=52. X is made by mixing A:B:C in ratio 2:2:6. Y is made by mixing A:B:C in ratio 10:2:2. Profit on X is thrice the profit on Y. 2. **Define variables:** Let $x$ = quantity of brand X produced. Let $y$ = quantity of brand Y produced. 3. **Component constraints:** For A: $2x + 10y \leq 50$ For B: $2x + 2y \leq 40$ For C: $6x + 2y \leq 52$ 4. **Profit function:** Let profit per unit of Y be $p$. Then profit per unit of X is $3p$. Objective function to maximize total profit: $$Z = 3p x + p y = p(3x + y)$$ Since $p$ is positive, maximize $3x + y$. 5. **Objective function:** $$\text{Maximize } Z = 3x + y$$ 6. **Solve graphically:** Plot constraints: - $2x + 10y \leq 50$ - $2x + 2y \leq 40$ - $6x + 2y \leq 52$ - $x \geq 0, y \geq 0$ Find intersection points of constraints: - Intersection of $2x + 10y = 50$ and $2x + 2y = 40$: Subtract second from first: $(2x + 10y) - (2x + 2y) = 50 - 40 \Rightarrow 8y = 10 \Rightarrow y = \frac{10}{8} = 1.25$ Substitute $y=1.25$ into $2x + 2(1.25) = 40 \Rightarrow 2x + 2.5 = 40 \Rightarrow 2x = 37.5 \Rightarrow x = 18.75$ - Intersection of $2x + 10y = 50$ and $6x + 2y = 52$: Multiply first by 1 and second by 5: $2x + 10y = 50$ $30x + 10y = 260$ Subtract first from second: $28x = 210 \Rightarrow x = 7.5$ Substitute $x=7.5$ into $2(7.5) + 10y = 50 \Rightarrow 15 + 10y = 50 \Rightarrow 10y = 35 \Rightarrow y = 3.5$ - Intersection of $2x + 2y = 40$ and $6x + 2y = 52$: Subtract first from second: $(6x + 2y) - (2x + 2y) = 52 - 40 \Rightarrow 4x = 12 \Rightarrow x = 3$ Substitute $x=3$ into $2(3) + 2y = 40 \Rightarrow 6 + 2y = 40 \Rightarrow 2y = 34 \Rightarrow y = 17$ 7. **Evaluate objective function $Z=3x + y$ at vertices:** - At $(0,0)$: $Z=0$ - At $(0,5)$ from $2x + 10y = 50$: $Z=3(0)+5=5$ - At $(18.75,1.25)$: $Z=3(18.75)+1.25=56.25+1.25=57.5$ - At $(7.5,3.5)$: $Z=3(7.5)+3.5=22.5+3.5=26$ - At $(3,17)$: $Z=3(3)+17=9+17=26$ 8. **Maximum profit at $(18.75,1.25)$** **Final answer:** Maximum profit is achieved by producing $x=18.75$ units of brand X and $y=1.25$ units of brand Y.