The dietitian's dependent nutrient system
A hospital dietitian is designing a daily supplement mix from three products, A, B and C. Per 100 g, A supplies 10 g protein, 8 g fat and 30 g carbohydrate; B supplies 20 g protein, 16 g fat and 60 g carbohydrate; C supplies 12 g protein, 6 g fat and 40 g carbohydrate. The target for a patient is 32 g protein, 22 g fat and 100 g carbohydrate. Let x, y and z be the amounts of A, B and C in hundreds of grams. A student writes the three balance equations, computes the determinant of the coefficient matrix, finds it is zero, and concludes that the target cannot be met because there is no unique solution. The dietitian needs a feasible nonnegative mix and also wants to know what happens if one product is unavailable. The student's conclusion is being challenged in a case-study seminar.
1. Show that the coefficient matrix is singular, but decide whether the target is still attainable. Solve the system and describe all nonnegative solutions.
Read the worked answer
Write x, y, z for hundreds of grams of A, B, C. The balance equations are:
10x+20y+12z=32 8x+16y+6z=22 30x+60y+40z=100.
The columns are A=(10,8,30), B=(20,16,60), C=(12,6,40). Since B=2A, the columns are linearly dependent, so the determinant is 0. That rules out a unique solution, but it does not rule out solutions.
Let u=x+2y. Then the mixture is uA+zC, so
10u+12z=32, 8u+6z=22, 30u+40z=100.
Multiply the second equation by 2: 16u+12z=44. Subtract the first: 6u=12, so u=2. Then 10(2)+12z=32 gives z=1. Check the third: 30(2)+40(1)=100, so the system is consistent.
Thus x+2y=2 and z=1, with x>=0, y>=0. There are infinitely many nonnegative solutions, for example (x,y,z)=(2,0,1), (0,1,1), or (1,0.5,1). The student's conclusion is wrong: a zero determinant rules out uniqueness, not existence.
2. If product C is unavailable, can the target be met exactly? Justify your answer with equations, not just a determinant.
Read the worked answer
With C unavailable, z=0. Then the target must be a multiple of A: (32,22,100)=u(10,8,30). From protein, u=3.2; from fat, u=22/8=2.75; these disagree. Equivalently, 8(3.2)=25.6, not 22. So the target cannot be met exactly without C. A determinant argument alone would not show this; the inconsistency comes from comparing the target vector with the span of A and B, and B is only 2A.
3. The dietitian wants equal amounts of A and B (x=y). Find the mix, or show it is impossible.
Read the worked answer
Set x=y. Then u=x+2y=3x. From the general solution u=2, so 3x=2, giving x=2/3. Thus y=2/3 and z=1. The mix is about 66.7 g of A, 66.7 g of B, and 100 g of C. It is nonnegative and meets the target because it is a special case of x+2y=2, z=1.