4. A registered nurse is trying to develop a diet plan for patients. The require

Question

4. A registered nurse is trying to develop a diet plan for patients. The required nutritional elements are the total in table 2: daily requirements of each nutritional element are as indicated Table 2 (Required Nutritional Element Total and Daily Requirements) Calories Not more than 2,700 calories Carbohydrates Not less than 300 grams Protein Not less than 250 grams Vitamins Not less than 60 units units of nutritional elements per unit of The nurse has four basic types to use when planning the menus. The food type are shown in the tabl le below. Note than the cost associated with a unit of ingredient also appears at the bottom of table 3 Table 3 (Required Nutritional Element and Units of Nutritional Elements Per Unit of Food Type) Vegetables Milk 160 Chicken ea 120 110 90 75 eme 210 130 190 50 150 120 130 70 Calories Protein Vitamins Carbohydrates 110 90 50 Cost per unit $0.42 S0.68 $0.32 $0.17 Moreover, due to dietary restrictions, the following aspects should also be considered when the developing the diet plan: The chicken food type should contribute at most 25% of the total calories intake that will result from the diet plan. ii) The vegetable food type should provide at least 30% of the minimum daily requirements for vitamins Provide a linear programming formulation for the above case. (No need to solve the problem.)

Explanation / Answer

SOLUTION:

(i). Let the unit of nutritional element of different element in chicken food type containing calaries, carbohydrate, protein and vitamin be x,y,z,p.

Quantity Analysis:

x<=25%(x+y+z+p)

x<=(25/100)(x+y+z+p)

4x<=(x+y+z+p)

3x<=y+z+p .........(1a)

x<=2700 ...........(1b)

y>=300 .............(1c)

z>=250 .............(1d)

p>=60 .............(1e)

Total cost=0.42x+0.68y+0.32z+0.17p .......(1f)

(ii).

Let the unit of nutritional element of different element in Vegetable food type containing calaries, carbohydrate, protein and vitamin be x,y,z,p.

Quantity Analysis:

p>=30%(x+y+z+p)

p>=(30/100)(x+y+z+p)

10p>=3x+3y+3z+3p

7p>=3x+3y+3z ..........(2a)

x<=2700 ...........(2b)

y>=300 .............(2c)

z>=250 .............(2d)

p>=60 .............(2e)

Total cost=0.42x+0.68y+0.32z+0.17p .......(2f)

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