A dietician is planning a meal that supplies certain quantities of vitamin C, calcium, and magnesium.
Three foods will be used, their quantities measured in milligrams. The nutrients supplied by these foods and the dietary requirements are given in the table below.
Nutrient Food 1 Food 2 Food 3 Total Required (mg)
Vitamin C 50 100 100 10025
Calcium 30 80 70 7145
Magnesium 20 55 60 5345
What quantity (mg) of Food 1 is necessary to meet the dietary requirements?
What quantity (mg) of Food 2 is necessary to meet the dietary requirements?
What quantity (mg) of Food 3 is necessary to meet the dietary requirements?
To find the quantity of each food necessary to meet the dietary requirements, we can set up a system of equations and solve it using the method of substitution.
Let's assume the quantities of Food 1, Food 2, and Food 3 are x, y, and z (in milligrams) respectively.
The first equation represents the requirement for Vitamin C:
50x + 100y + 100z = 10025
The second equation represents the requirement for Calcium:
30x + 80y + 70z = 7145
The third equation represents the requirement for Magnesium:
20x + 55y + 60z = 5345
We can solve this system of equations to find the values of x, y, and z.
To do this, let's solve the first equation for x:
50x = 10025 - 100y - 100z
x = (10025 - 100y - 100z)/50
x = (200.5 - 2y - 2z)
Next, let's substitute x in the second and third equations:
30(200.5 - 2y - 2z) + 80y + 70z = 7145
(6015 - 60y - 60z) + 80y + 70z = 7145
20y + 10z = 1130 ............(Equation 1)
20(200.5 - 2y - 2z) + 55y + 60z = 5345
(4010 - 40y - 40z) + 55y + 60z = 5345
15y + 20z = 1335 ............(Equation 2)
Now, we have a system of two equations with two unknowns (y and z). We can solve this system of equations to find the values of y and z.
Multiplying Equation 1 by 2 and Equation 2 by 3:
40y + 20z = 2260 ............(Equation 3)
45y + 60z = 4005 ............(Equation 4)
Subtracting Equation 3 from Equation 4:
(45y + 60z) - (40y + 20z) = 4005 - 2260
5y + 40z = 1745
5(y + 8z) = 1745
y + 8z = 349 ............(Equation 5)
Now, we have a system of two equations with two unknowns (y and z) again.
Let's solve Equation 5 for y:
y = 349 - 8z
Now, substitute this value of y in Equation 1:
20(349 - 8z) + 10z = 1130
6980 - 160z + 10z = 1130
-150z = -5850
z = 39
Substitute the value of z in Equation 5:
y + 8(39) = 349
y + 312 = 349
y = 37
We have found the values of y and z. Now, substitute these values back into the equation for x:
x = (200.5 - 2y - 2z)
x = (200.5 - 2(37) - 2(39))
x = (200.5 - 74 - 78)
x = 48.5
Therefore, the quantities necessary to meet the dietary requirements are:
Food 1: 48.5 mg
Food 2: 37 mg
Food 3: 39 mg