Question

Bob, a nutritionist who works for the university medical center, has been asked to prepare special diets for two patients, susan and tom. bob has decided that susan's meals should contain at least 320 mg of calcium, 17 mg of iron, and 42 mg of vitamin c, whereas tom's meals should contain at least 280 mg of calcium, 13 mg of iron, and 34 mg of vitamin
c. bob has also decided that the meals are to be prepared from three basic food

Answer 1

Answer and Explanation:

Possibly, you forgot to attach the table with the contents of each food. Therefore, I calculated the task with similar values.

1. Assume you have Food A, B, and C. Each has certain amount of Calcium, Iron, and Vit C.

            Let's designate: Calcium → X oz , Iron → Y oz, vit C → Z oz

2. Assume you have following amount of Calcium in each food:

• Food A = 30 mg/oz
• Food B = 25 mg/oz
• Food C = 20 mg/oz

3. Assume you have following amount of Iron in each food:

• Food A = 1 mg/oz
• Food B = 1 mg/oz
• Food C = 2 mg/oz

4. Assume you have following amount of vit C in each food:

• Food A = 2 mg/oz
• Food B = 5 mg/oz
• Food C = 4 mg/oz

Calculations:

→ You have 3 different food and each has Calcium in it. You need exactly 320mg of Calcium per day. Therefore, each food would comprise certain proportion of daily Calcium intake. We can come up with equation: 30X+25Y+20Z = 390 mg

• Similarly, for iron and vitamin C, the next equations could be used:

        X+Y+2Z=17 - Iron

        2X+5Y+4Z=42 - Vit C

→ Now you have a system of equations:

30X+25Y+20Z = 390 (1)

X+Y+2Z=17 (2)

2X+5Y+4Z=42 (3)

→ Look for similarities! (1) and (3) seem to be similar. We can multiply the (3) by -5 to obtain: -10X-25Y-20Z=-210. We add the (1) and modified (3) equations.

30X+25Y+20Z = 390

-10X-25Y-20Z=-210

__________________

20X+0+0=180

20X=180

→ X=9 (oz of Calcium)

→ Now, we can put x=9 into equations above to ease our job:

270+25Y+20Z=390 → 25Y+20Z=390-270=120

9+Y+Z=17 → Y+2Z=17-9=8

18+5Y+4Z=42 → 5Y+4Z=42-18=24

→ Similarly, we can use Eq-s (2) and (3). We can multiply the (2) by -5 to obtain: -5Y-10Z=-40. We add the (3) and modified (2) equations.

-5Y-10Z=-40

5Y+4Z=24

__________________

-6Z=-24

Z=4

→ Z=4 (oz of vit C)

→ Now, we can take X+Y+Z=17 (2) and just put in known values(X=9 and Z=4) to obtain Y:

9+Y+4=17

→ Y=17-9-4=4 (oz of Iron)

That's all! So, if you have table values, you can just re-write the equations and doing it stepwise, as it is shown here, you can manage to calculate every value you need :)