Final answer:
To achieve the desired diet of 25 grams protein and 7 grams fat, the technician should mix 100 grams of Mix A with 50 grams of Mix B. This is solved using a system of linear equations.
Step-by-step explanation:
To determine how many grams of each food mix are needed to satisfy the dietary requirements for one animal, we can set up a system of linear equations based on the protein and fat content of each mix.
Let x be the grams of Mix A and y be the grams of Mix B. The protein content in Mix A (20%) and Mix B (50%) can be represented as 0.20x and 0.50y, respectively. Similarly, the fat content in Mix A (8%) and Mix B (5%) can be represented as 0.08x and 0.05y, respectively. The requirements are 25 grams of protein and 7 grams of fat.
We now have the following system of equations:
- 0.20x + 0.50y = 25 (protein equation)
- 0.08x + 0.05y = 7 (fat equation)
To solve this system, we can use the method of substitution or elimination. If we multiply the first equation by 5 and the second equation by 100, we obtain two new equations that make it easier to eliminate variables. We get:
- 1x + 2.5y = 125
- 8x + 5y = 700
To eliminate y, we can multiply the first equation by 2 and the second equation by 1, then subtract the second equation from the first, which gives us a new equation in terms of x only. We find that x equals 100 grams. We can then substitute this value back into one of the original equations to find y, which equals 50 grams.
Therefore, to create a diet with the correct nutritional content of 25 grams of protein and 7 grams of fat for one animal, the technician should use 100 grams of Mix A and 50 grams of Mix B