An animal breeder can buy four types of food for Vietnamese pot-bellied pigs. Each case of Brand A contains 25 units of fiber, 40 units of protein, and 40 units of fat. Each case of Brand B contains 100 units of fiber, 130 units of protein, and 110 units of fat. Each case of Brand C contains 275 units of fiber, 350 units of protein, and 310 units of fat. Each case of Brand D contains 200 units of fiber, 260 units of protein, and 200 units of fat. How many cases of each brand should the breeder mix together to obtain a food that provides 3975 units of fiber, 5100 units of protein, and 4440 units of fat? Brand B, z represent the number of cases of Brand C, and $$ w $$ represent be the number of cases of Brand D. There are four ways in which the breeder can mix brands to obtain a food that provides 3975 units of fiber, 5100 units of protein, and 4440 units of fat. If $$ w=0 $$, the solution is $$ (0,15,9,0) $$. If $$ w=1 $$, the solution is $$ (1,10,10,1) $$. If $$ w=2 $$, the solution is $$ (\square, \square, \square, 2) $$.

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We want to mix four brands (A, B, C, and D) to get a total of 3975 units of fiber, 5100 units of protein, and 4440 units of fat. The food compositions by case are:

• Brand A: 25 fiber, 40 protein, 40 fat  
 • Brand B: 100 fiber, 130 protein, 110 fat  
 • Brand C: 275 fiber, 350 protein, 310 fat  
 • Brand D: 200 fiber, 260 protein, 200 fat

Let x, y, z, and w denote the number of cases of Brands A, B, C, and D respectively.

Thus we have three equations:

1. Fiber:  25x + 100y + 275z + 200w = 3975  
 2. Protein: 40x + 130y + 350z + 260w = 5100  
 3. Fat:   40x + 110y + 310z + 200w = 4440

We are told that there are four ways to get the mix. Two examples are given:
 • When w = 0, the solution is (x, y, z, w) = (0, 15, 9, 0)  
 • When w = 1, the solution is (x, y, z, w) = (1, 10, 10, 1)

We now wish to find the solution when w = 2.

Step 1. Subtract the nutrients provided by 2 cases of Brand D:
 For 2 cases, Brand D contributes:
  Fiber: 200 × 2 = 400  
  Protein: 260 × 2 = 520  
  Fat: 200 × 2 = 400

Subtract these from the totals:
 Fiber left: 3975 − 400 = 3575  
 Protein left: 5100 − 520 = 4580  
 Fat left:  4440 − 400 = 4040

So we now have three equations in x, y, and z:

a) 25x + 100y + 275z = 3575  
 b) 40
To achieve the desired nutrient totals by mixing Brands A, B, C, and D, the breeder can use the following combinations when $$ w = 2 $$: - **Solution when $$ w = 0 $$:** (0, 15, 9, 0) - **Solution when $$ w = 1 $$:** (1, 10, 10, 1) - **Solution when $$ w = 2 $$:** (2, 5, 11, 2) These combinations provide the required 3975 units of fiber, 5100 units of protein, and 4440 units of fat.

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