Question 1148051
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            This problem can be solved  VERY  QUICKLY,  by the very simple way and without using the Linear programming approach.



<pre>
Let  x = the amount of beef, in kilograms;
     y = the amount of pork;
     z = the amount of fat.


Your constraint equations are:

     x + y + z = 50
     z <= 10
     y <= 0.25x
     x >= 0, y >= 0


Your objective function to minimize is

     cost = 6x + 4y + 2z


Since fat is cheapest ingredient, the butcher decides to use all 10 kg of fat.
Than your constraints take the form

     x + y = 40          (1)
     y <= 0.25x          (2)
     x >= 0, y >= 0      (3)


and your task now is to minimize the form

     cost = 6x + 4y + 20

which is the same as to minimize

     cost' = 6x + 4y.    (4)


So, your problem is just reduced to 2 variables.


Next, from (1), express  y = 40-x  and substitute it into (4) and (2). You will get the function to minimize

     cost' = 6x + 4*(40-x) = 160 + 2x    (5)

under the restriction (2)

    40-x <= 0.25x,  or  40 <= x + 0.25x = 1.25x, which is  x >= {{{40/1.25}}} = 32.   (6)


Formula (5) says "than lesser is x, than lesser is the value of the function c' = 160+2x."


But formula (6) says that you CAN NOT take x lesser than 32.


Thus,  x= 32 kilograms of beef is the solution.


Then the full and complete <U>ANSWER</U> is


    x= 32 kilograms of beef,  y= 40-32 = 8 kilograms of pork and z= 10 kilograms of fat.
</pre>


Solved (!)


Solved even without using Linear programming approach &nbsp;(!) &nbsp;(!)


Solved at the level accessible even for &nbsp;6 - 7- 8 &nbsp;grade middle school students level &nbsp;(!) &nbsp;(!!) &nbsp;(!!!)