1. **Stating the problem:** We have 6 kg of flour and 5 kg of butter to make two types of cakes, A and B. Each cake A requires 200 g flour and 200 g butter. Each cake B requires 200 g flour and 150 g butter. Profit per cake A is 75000 and per cake B is 60000. We want to find the number of cakes A and B to maximize profit. 2. **Convert units:** 6 kg = 6000 g flour, 5 kg = 5000 g butter. 3. **Define variables:** Let $x$ = number of cake A, $y$ = number of cake B. 4. **Constraints:** Flour: $200x + 200y \leq 6000$ Butter: $200x + 150y \leq 5000$ Also, $x \geq 0$, $y \geq 0$. 5. **Profit function:** $$P = 75000x + 60000y$$ 6. **Simplify constraints:** Divide flour constraint by 200: $$x + y \leq \cancel{\frac{6000}{200}}^{30}$$ Divide butter constraint by 50: $$4x + 3y \leq \cancel{\frac{5000}{50}}^{100}$$ 7. **Find feasible region vertices:** - Intersection of $x + y = 30$ and $4x + 3y = 100$: Multiply first by 3: $3x + 3y = 90$ Subtract from second: $(4x + 3y) - (3x + 3y) = 100 - 90 \Rightarrow x = 10$ Then $y = 30 - 10 = 20$ - Intercepts: For $x + y \leq 30$: $x=0 \Rightarrow y=30$, $y=0 \Rightarrow x=30$ For $4x + 3y \leq 100$: $x=0 \Rightarrow y=\frac{100}{3} \approx 33.33$ $y=0 \Rightarrow x=25$ 8. **Evaluate profit at vertices:** - At $(0,0)$: $P=0$ - At $(0,30)$: $P=75000(0)+60000(30)=1800000$ - At $(25,0)$: $P=75000(25)+60000(0)=1875000$ - At $(10,20)$: $P=75000(10)+60000(20)=750000+1200000=1950000$ 9. **Maximum profit:** $1950000$ at $x=10$, $y=20$. **Final answer:** The maximum profit is 1950000, which corresponds to option C.