SOLUTION: One gram of soybean meal provides at least 2.5 units of vitamins and 5 calories. One gram of meat byproducts provides at least 4.5 units of vitamins and 3 calories. One gram of gra

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Question 1142297: One gram of soybean meal provides at least 2.5 units of vitamins and 5 calories. One gram of meat byproducts provides at least 4.5 units of vitamins and 3 calories. One gram of grain provides at least 5 units of vitamins and 10 calories. If a gram of soybean meal costs 6 ​cents, a gram of meat byproducts 7 ​cents, and a gram of grain 9 ​cents, what mixture of these three ingredients will provide at least 48 units of vitamins and 54 calories per serving at minimum​ cost? What will be the minimum​ cost?
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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Let X = the mass of soybean meal consumed (in grams);

    Y = the mass of meat;

    Z = the mass of grain.


The objective function to minimize is the cost


    C(X,Y,Z) = 6*X + 7*Y + 9*Z  cents.      (1)


The constraints are


    2.5*X + 4.5*Y + 5*Z  >= 48  units of vitamins,    (2)

    
    5*X   + 3*Y   + 10*Z >= 54  calories.             (3)


Other constraints are  X >= 0;  Y>= 0,  and  Z >= 0.  (4)


Now, a remarkable fact is that the solution to this minimax problem can be obtained ANALYTICALLY.


Constraints (2) and (3) represent two planes in 3D:


    2.5*X + 4.5*Y + 5*Z   = 48                        (5)

    5*X   + 3*Y   + 10*Z  = 54                        (6)


These planes are not parallel -- hence, their intersection is a straight line.

The idea is to present this straight line in a parametric form - then the solution of the minimax problem
on this straight line will be easy.


Multiply equation (5) by 2 (both sides) and then subtract equation (6) from the obtained equation.
You will get


    3Y - 9Y = 54 - 2*48,   or     -6Y = -42.  

Hence,  

    Y = 7.                                            (7)


Thus we found that the intersection of two planes (5) and (6) is a straight line, which lies on the plane Y = 7.

Subctitute Y =7 into equations (5) and (6). You will get then


    2.5*X + 4.5*7 + 5Z  = 48            (5')

    5*X   + 3*7   + 10Z = 54,           (6')


or, collecting all constant terms on the right side


    2.5*X + 5Z   = 16.5,                (5'')

    5*X   + 10*Z = 33,                  (6'')


Equations (5'') and (6'') are DEPENDENT (which is OBVIOUS).

Hence, two equations (5'') and (6'') represent THE SAME plane.


So, our straight line is the intersection of planes (7) and (6'').


Now, from equation (6''),

     X = 6.6 - 2Z.


Thus our stright line in parametric form is


    X = 6.6 - 2Z,  Y = 7.               (8)


Substitute (8) into the objective function (1). You will get

    C(X,Y,Z) = 6*X + 7*Y + 9*Z = 6*(6.6 - 2Z) + 7*7 + 9*Z = 39.6 - 12Z + 49 + 9Z = -3Z + 88.6.    (9)


Thus, on our line the objective function is presented as the linear function (9) of Z.

We see that when Z increases from 0 to positive values, the function  (9) decreases.
But Z can increase only till  X = 6.6 - 2Z  is >= 0    (is non-negative).
Hence, the linear function (9) has the minimum at Z = 6.6/2 = 3.3.

Then X = 6.6 - 2*3.3 = 0, according to (8).


Thus we just obtained the solution to our minimax problem:

    The minimum solution point is X= 0;  Y= 7  and  Z= 3.3 

    and the mimimum cost is  -3*Z + 88.6 = -3*3.3 + 88.6 =  78.7.


ANSWER.  The minimum cost is 78.70 cents and it is achieved at this diet:

         0 gram of soybean meal;  7 gram of meat,  and 3.3 gram of grain.

Solved.