a dietitian in a hospital is to arrange a special dietusing three basic foods. T

Question

a dietitian in a hospital is to arrange a special dietusing three basic foods. The diet i sto include exactly 340 unitsof calcium, 180 units of iron and 220 units of vitamin A Thenumber of units per ounce of each special ingredient for each ofthe foods is indicated in the table.   please showhow              Food a        foodb     food c cal            30             10          20 iron           10             10           20 vita            10             30           20 a how many ounces of each food must be used tomeet the diet requirements b   How is the diet in part a affected if food c isnot used c    How is the diet in part a affected if thevitamin a requirement is dropped a dietitian in a hospital is to arrange a special dietusing three basic foods. The diet i sto include exactly 340 unitsof calcium, 180 units of iron and 220 units of vitamin A Thenumber of units per ounce of each special ingredient for each ofthe foods is indicated in the table.   please showhow              Food a        foodb     food c cal            30             10          20 iron           10             10           20 vita            10             30           20 a how many ounces of each food must be used tomeet the diet requirements b   How is the diet in part a affected if food c isnot used c    How is the diet in part a affected if thevitamin a requirement is dropped

Explanation / Answer

. Part A: 1.   This is a problem of three unknowns... soyou'll need three independent equations to solve for a, b, andc; fortunately, we have enough data to create the threeequations. . 2.   30a + 10b + 20c =340      ;   this isthe combination that will yield 340 units of calcium 3.   10a + 10b + 20c =180      ;   this isthe combination that will yield 180 units of iron 4.   10a + 30b + 20c =220      ;   this is thecombination that will yield 220 units of vitamin A . 5.   Now that we have three equations, we solve forthe unknowns a, b, and c. . 6.1.   multiply '3' by -1 and add to '2': 6.2.   (-10a - 10b - 20c = -180) + (30a +10b + 20c = 340) ==> 20a + 0b + 0c = 160   a =8 . 7.1.   multiply '3' by -1 and add to '3': 7.2.   (-10a - 10b - 20c = -180) + (10a +30b + 20c = 220) ==> 0a + 20b + 0c = 40   b =2 . 8.1.   Now that we have two of the unknowns solved,plug the values for a and b into any one of the original equationsand solve for c.   (I'll use the equation givenat '3'.) 8.2.   (30)*(8) + (10)*(2) + 20c = 340 8.3.   240 + 20 + 20c = 340 8.4.   20c = 340 - 240 - 20 8.5.   20c = 80 8.6.   c =4 . ============================================================== . Part B: 09.   In part_B, I will not be able to solve for thequantity of foods to achieve the specified solution; I can solvefor any two, but the third item will be dependent upon the solutionfor the other two. For example, suppose I chose to solve foriron and vitamin_A, my equations are: 10.   10a + 10b =180      ;   this isthe combination that will yield 180 units of iron 12.   10a + 30b =220      ;   this is thecombination that will yield 220 units of vitamin A . 13.   multiply '10' by -1 and add to '12': 14.   (-10a - 10b = -180) + (10a + 30b = 220)==> 0a + 20b + 0c = 40   b = 2 15.   substitute the result in '14' into '10' andsolve for 'a': 16.   10a + (10)*(2) = 180  a = 16 17.   This combination of a and b yields the properamount for iron and vitamin_A, but I'm stuck with the result forcalcium which is: (30)*(16) + (10)*(2) = 500 units ofcalcium. You might try the other two combinations and thenpick the solution that is closest to the originalprescription. . ============================================================== . Part C: 18.   In part_C, I will have more food combinationsthan needed to meet the calcium and iron needs; this will result inone need being a choice between the remaining two fooditems. The two equations are: 19.   30a + 10b + 20c =340      ;   this isthe combination that will yield 340 units of calcium 20.   10a + 10b + 20c =180      ;   this isthe combination that will yield 180 units of iron . 21.   multiply '20' by -1 and add to '19': 22.   (-10a - 10b - 20c = -180) + (30a +10b + 20c = 340) ==> 20a + 0b + 0c = 160   a =8 23.   substitute the value for 'a' into '20' andsolve for b and c. 24.   (10)*(8) + 10b + 20c = 180  ==>   10b + 20c = 100  ;   divide through by 10 25.   b + 2c =10         b= 10 -2c         or       c= (10 - b)/2 26.1.   Thus, I can either chose the quantity of cand then calculate b...     or... 26.2.   chose the quantity of b and calculatethe amount of c. .      12.   10a + 30b =220      ;   this is thecombination that will yield 220 units of vitamin A . 13.   multiply '10' by -1 and add to '12': 14.   (-10a - 10b = -180) + (10a + 30b = 220)==> 0a + 20b + 0c = 40   b = 2 15.   substitute the result in '14' into '10' andsolve for 'a': 16.   10a + (10)*(2) = 180  a = 16 17.   This combination of a and b yields the properamount for iron and vitamin_A, but I'm stuck with the result forcalcium which is: (30)*(16) + (10)*(2) = 500 units ofcalcium. You might try the other two combinations and thenpick the solution that is closest to the originalprescription. . ============================================================== . Part C: 18.   In part_C, I will have more food combinationsthan needed to meet the calcium and iron needs; this will result inone need being a choice between the remaining two fooditems. The two equations are: 19.   30a + 10b + 20c =340      ;   this isthe combination that will yield 340 units of calcium 20.   10a + 10b + 20c =180      ;   this isthe combination that will yield 180 units of iron . 21.   multiply '20' by -1 and add to '19': 22.   (-10a - 10b - 20c = -180) + (30a +10b + 20c = 340) ==> 20a + 0b + 0c = 160   a =8 23.   substitute the value for 'a' into '20' andsolve for b and c. 24.   (10)*(8) + 10b + 20c = 180  ==>   10b + 20c = 100  ;   divide through by 10 25.   b + 2c =10         b= 10 -2c         or       c= (10 - b)/2 26.1.   Thus, I can either chose the quantity of cand then calculate b...     or... 26.2.   chose the quantity of b and calculatethe amount of c. .      20.   10a + 10b + 20c =180      ;   this isthe combination that will yield 180 units of iron . 21.   multiply '20' by -1 and add to '19': 22.   (-10a - 10b - 20c = -180) + (30a +10b + 20c = 340) ==> 20a + 0b + 0c = 160   a =8 23.   substitute the value for 'a' into '20' andsolve for b and c. 24.   (10)*(8) + 10b + 20c = 180  ==>   10b + 20c = 100  ;   divide through by 10 25.   b + 2c =10         b= 10 -2c         or       c= (10 - b)/2 26.1.   Thus, I can either chose the quantity of cand then calculate b...     or... 26.2.   chose the quantity of b and calculatethe amount of c. .     

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