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? Let $$ x $$ represent the number of cases of Brand A, y represent the number of cases of 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 $$ (\square, \square, \square, 1) $$.

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We are given that each case of the food brands provides the following nutrients:

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

The breeder wants a mix providing exactly 3975 units of fiber, 5100 units of protein, and 4440 units of fat. Let x, y, z, and w be the number of cases of Brands A, B, C, and D, respectively. Then the nutrient equations are:

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

One of the ways to mix the foods (when w = 0) is given as (0, 15, 9, 0).

Now we need to find the solution when w = 1.

Step 1. Substitute w = 1 into the equations.

For fiber:
 25x + 100y + 275z + 200(1) = 3975  
 25x + 100y + 275z = 3975 − 200 = 3775           (1)

For protein:
 40x + 130y + 350z + 260(1) = 5100  
 40x + 130y + 350z = 5100 − 260 = 4840         (2)

For fat:
 40x + 110y + 310z + 200(1) = 4440  
 40x + 110y + 310z = 4440 − 200 = 4240         (3)

Step 2. Simplify the equations.

Divide (1) by 5:
 5x + 20y + 55z = 3775/5 = 755           (1a)

Divide (2) by 10:
 4x + 13y + 35z = 4840/10 = 484          (2a)

Divide (3) by 10:
 4x + 11y + 31z = 4240/10 = 424          (3a)

Step 3. Eliminate x by subtracting (3a) from (2a).

(4x + 13y + 35z) − (4x + 11y + 31z) = 484 − 424  
 (4x − 4x) + (13y − 11y) + (35z − 31z) = 60  
 2y + 4z = 60  
Divide by 2:
 y + 2z = 30  
Thus,
 y = 30 − 2z                (4)

Step 4. Substitute y from (4) into (3a) to solve for x.

Equation (3a) is:
 4x + 11y + 31z = 424

Replace y:
 4x + 11(30 − 2z) + 31z = 424  
 4x + 330 − 22z + 31z = 424  
 4x + 330 + 9z = 424  
Solve for x:
 4x = 424 − 330 − 9z = 94 − 9z  
 x = (94 − 9z)/4          (5)

x must be an integer, so (94 − 9z) must be divisible by 4. Notice that 9 mod 4 is 1. Thus the condition becomes:
 (94 − 9z) ≡ (94 − z) ≡ 0 (mod 4)

Since 94 mod 4 is 2, we require:
 2 − z ≡ 0 (mod 4)  
which means
 z ≡ 2 (mod 4)  
So z can be 2, 6, 10, etc. We now use (1a) to find the correct z.

Step 5. Substitute x and y from (5) and (4) into (1a).

Equation (1a):
 5x + 20y + 55z = 755

Substitute x and y:
 5[(94 − 9z)/4] + 20(30 − 2z) + 55z = 755

Multiply the entire equation by 4 to clear the fraction:
 5(94 − 9z) + 80(30 − 2z) + 220z = 755 × 4  
 5(94 − 9z) = 470 − 45z  
 80(30 − 2z) = 2400 − 160z  
So the equation becomes:
 (470 − 45z) + (2400 − 160z) + 220z = 3020

Combine like terms:
 470 + 2400 + (−45z − 160z + 220z) = 3020  
 2870 + 15z = 3020  
Subtract 2870 from both sides:
 15z = 3020 − 2870 = 150  
Solve:
 z = 150 / 15 = 10

Step 6. Now find y and x.

Using (4):
 y = 30 − 2z = 30 − 2(10) = 30 − 20 = 10

Using (5):
 x = (94 − 9(10)) / 4 = (94 − 90)/4 = 4/4 = 1

Thus, when w = 1 the solution is:
 x = 1, y = 10, z = 10, w = 1

Step 7. Verify the solution.

Fiber:
 25(1) + 100(10) + 275(10) + 200(1) = 25 + 1000 + 2750 + 200 = 3975

Protein:
 40(1) + 130(10) + 350(10) + 260(1) = 40 + 1300 + 3500 + 260 = 5100

Fat:
 40(1) + 110(10) + 310(10) + 200(1) = 40 + 1100 + 3100 + 200 = 4440

The solution checks out.

Thus, if w = 1, the case mix is:
 (1, 10, 10, 1)
When $$ w = 1 $$, the breeder should use 1 case of Brand A, 10 cases of Brand B, 10 cases of Brand C, and 1 case of Brand D to achieve the desired nutrient levels.

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