Find how many ounces of each type of food should be used in a meal so that the m

Question

Find how many ounces of each type of food should be used in a meal so that the minimum requirements of calcium, iron and vitamin C are met for each patient? I will rate!!! I will rate!!! Bob a nutritionist who works for the University Medical Center, has been asked to prepare diets for two patients, Susan and Tom. Bob has decided that Susan's meals should contain at least 400mg of calcium, 20mg of iron and 50mg of Vitamin C. Whereas Tom's meals should contain at least 350mg of calcium, 15mg of iron, and 40mg of vitamin C. Bob has also decided that the meals are to be prepared in three basic food groups:food A, food B, and food C. The special nutritional contents of these foods are summarized in the accompanying table. Find how many ounces of each type of food should be used in a meal so that the minimum requirements of calcium, iron and vitamin C are met for each patient?

Explanation / Answer

Bob a nutritionist who works for the University Medical Center, has been asked to prepare diets for two patients, Susan and Tom. Bob has decided that Susan's meals should contain at least 400mg of calcium, 20mg of iron and 50mg of Vitamin C. Whereas Tom's meals should contain at least 350mg of calcium, 15mg of iron, and 40mg of vitamin C.

Bob has also decided that the meals are to be prepared in three basic food groups:food A, food B, and food C. The special nutritional contents of these foods are summarized in the accompanying table.


Find how many ounces of each type of food should be used in a meal so that the minimum requirements of calcium, iron and vitamin C are met for each patient?


The matrices were
A 30 1 2
B 25 1 5
C 20 2 4
for Calcium iron and vitamin C respectively

All we have to do is set up a system of equations for each patient. We distribute the letters of the food to each vitamin and then group them according to vitamin and set them equal to the recommended amount to see how much they will need in each group to reach their recommended amount

30A 1A 2A
25B 1B 5B
20C 2C 4C

Now let's group them, Bob has decided that Susan's meals should contain at least 400mg of calcium, 20mg of iron and 50mg of Vitamin C

30A+25B+20C=400
1A+ 1B +2C =20
2A + 5B +4C =50

Now we can solve the set of systems using the elimination method. Since the bottom 2 are small numbers, let's eliminate something from there first.
Let's eliminate B
To do this we multiply the top by -5

-5(1A+ 1B +2C =20)
2A + 5B +4C =50

-5A -5B -10C =-100
2A + 5B +4C =50                     Now we add them together and get
-3A-6C=-50 Now let's eliminate B from the top one and one from the bottom
30A+25B+20C=400
1A+ 1B +2C =20 Multiply the bottom by -25     30A+25B+20C=400
-25(1A+ 1B +2C =20) 30A+25B+20C   =400
-25A+ -25B -50C =-500    Now add them together -5A-30C=-100 Now combine the 2 -3A-6C=-50 -5A-30C=-100 Lets multiply the top by -5 to eliminate the C -5(-3A-6C=-50)    -5A-30C=-100   15A+30C=250 -5A-30C=-100         Now add them together 10A=150 A=15 Now plug this in to one of the previous equations that only contained A and C -5A-30C=-100 -5(15)-30C=-100 -75-30C=-100 -30C=-25 C=5/6    -5A-30C=-100   15A+30C=250 -5A-30C=-100         Now add them together 10A=150 A=15 Now plug this in to one of the previous equations that only contained A and C -5A-30C=-100 -5(15)-30C=-100 -75-30C=-100 -30C=-25 C=5/6 -5A-30C=-100         Now add them together 10A=150 A=15 Now plug this in to one of the previous equations that only contained A and C -5A-30C=-100 -5(15)-30C=-100 -75-30C=-100 -30C=-25 C=5/6 Now plug into one equation to get B 2A + 5B +4C =50 30+5B+4(5/6)=50 30+5B+20/6=50 5B+20/6=20 5B+20/6=120/6 5B=120/6-20/6 5B=100/6 B=100/30 B=10/3 Now let's plug this into another equation to check our work 1A+ 1B +2C =20 15+10/3+2(5/6)=20 15+20/6+10/6=20 15+30/6=20 15+5=20 Our anwers our correct Now let's calculate how much is required for Tom. Tom's meals should contain at least 350mg of calcium, 15mg of iron, and 40mg of vitamin C.
30A+25B+20C=350
1A+ 1B +2C =15 2A + 5B +4C =40 Let's eliminate B -5(1A+ 1B +2C =15)    2A + 5B +4C =40 -5A-5B-10C=-75 2A +5B+4C =40 -3A-6C=-35     30A+25B+20C=350
-25(1A+ 1B +2C =15) 30A+25b+20C=350 -25A-25B-50C=-375 -5A -30C = -25 Combine them -3A-6C=-35 -5A-30C=-25 Get rid of C -5(-3A-6C=-35) -5A-30C=-25 15A+30C=175 -5A-30C=-25 10A=150 A=15 -5(15)-30C=-25 -75-30C=-25 -30C=50 C=-5/3 1A+ 1B +2C =15 15+B-10/3=15 B=10/3 10A=150 A=15 -5(15)-30C=-25 -75-30C=-25 -30C=50 C=-5/3 1A+ 1B +2C =15 15+B-10/3=15 B=10/3

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