1. **Problem statement:** We want to find if a mixture of components A, B, and C can produce a food with given percentages of protein (Eiweiß), carbohydrates (Kohlenhydrate), and fat (Fett). 2. **Given data:** - Component A: 30% protein, 30% carbohydrates, 40% fat - Component B: 50% protein, 30% carbohydrates, 20% fat - Component C: 20% protein, 70% carbohydrates, 10% fat 3. **Variables:** Let $x$, $y$, and $z$ be the proportions of A, B, and C in the mixture, respectively. 4. **Constraints:** The proportions must sum to 1: $$x + y + z = 1$$ 5. **a) Target mixture:** 47% protein, 35% carbohydrates, 18% fat. Set up the system: $$\begin{cases} 30x + 50y + 20z = 47 \\ 30x + 30y + 70z = 35 \\ 40x + 20y + 10z = 18 \\ x + y + z = 1 \end{cases}$$ 6. **Solve the system:** From the sum equation: $$z = 1 - x - y$$ Substitute into the first three equations: $$30x + 50y + 20(1 - x - y) = 47$$ $$30x + 30y + 70(1 - x - y) = 35$$ $$40x + 20y + 10(1 - x - y) = 18$$ Simplify each: $$30x + 50y + 20 - 20x - 20y = 47 \Rightarrow 10x + 30y = 27$$ $$30x + 30y + 70 - 70x - 70y = 35 \Rightarrow -40x - 40y = -35 \Rightarrow 40x + 40y = 35$$ $$40x + 20y + 10 - 10x - 10y = 18 \Rightarrow 30x + 10y = 8$$ 7. **Simplify equations:** $$10x + 30y = 27$$ $$40x + 40y = 35$$ $$30x + 10y = 8$$ Divide second equation by 5: $$8x + 8y = 7$$ 8. **Solve for $x$ and $y$:** From first equation: $$10x + 30y = 27$$ From third equation: $$30x + 10y = 8$$ Multiply first by 3: $$30x + 90y = 81$$ Subtract third equation: $$(30x + 90y) - (30x + 10y) = 81 - 8 \Rightarrow 80y = 73 \Rightarrow y = \frac{73}{80} = 0.9125$$ Substitute $y$ into first equation: $$10x + 30(0.9125) = 27 \Rightarrow 10x + 27.375 = 27 \Rightarrow 10x = -0.375 \Rightarrow x = -0.0375$$ 9. **Find $z$:** $$z = 1 - x - y = 1 - (-0.0375) - 0.9125 = 0.125$$ 10. **Interpretation:** $x$ is negative, which is not possible for proportions. Therefore, no valid mixture exists for part (a). --- 11. **b) Target mixture:** 40% protein, 40% carbohydrates, fat unspecified. Set up the system with only protein and carbohydrates: $$\begin{cases} 30x + 50y + 20z = 40 \\ 30x + 30y + 70z = 40 \\ x + y + z = 1 \end{cases}$$ 12. **Substitute $z = 1 - x - y$:** $$30x + 50y + 20(1 - x - y) = 40 \Rightarrow 30x + 50y + 20 - 20x - 20y = 40 \Rightarrow 10x + 30y = 20$$ $$30x + 30y + 70(1 - x - y) = 40 \Rightarrow 30x + 30y + 70 - 70x - 70y = 40 \Rightarrow -40x - 40y = -30 \Rightarrow 40x + 40y = 30$$ 13. **Simplify:** $$10x + 30y = 20$$ $$40x + 40y = 30$$ Divide second by 10: $$4x + 4y = 3$$ 14. **Solve for $x$ and $y$:** Multiply first by 4: $$40x + 120y = 80$$ Multiply second by 10: $$40x + 40y = 30$$ Subtract second from first: $$(40x + 120y) - (40x + 40y) = 80 - 30 \Rightarrow 80y = 50 \Rightarrow y = \frac{50}{80} = 0.625$$ Substitute $y$ into $4x + 4y = 3$: $$4x + 4(0.625) = 3 \Rightarrow 4x + 2.5 = 3 \Rightarrow 4x = 0.5 \Rightarrow x = 0.125$$ 15. **Find $z$:** $$z = 1 - x - y = 1 - 0.125 - 0.625 = 0.25$$ 16. **Interpretation:** All proportions are positive and sum to 1, so a mixture exists for part (b). **Final answers:** - a) No valid mixture exists. - b) Mixture proportions: $x=0.125$, $y=0.625$, $z=0.25$.