An animal feed to be mixed

Question

An animal feed to be mixed from soybean meal and oats must contain at least 168 lb of protein, 24lb offat, and 14 lb of mineral ash. Each sack of soybeans costs $ 21 and contains 70 lb of protein, 8 lb of fat, and 7 lb of mineral ash. Each sack of oats costs $7 and contains 21 lb of protein,7 lb of fat, and 1 lb of mineral ash. How many sacks of each should be used to satisfy the minimum requirements at minimum cost?

Answer 1

Let x be the number of sacks of soybeans.
Ley y be the number of sacks of oats.

From the given data:

Protein: 70x + 21y = 168
21y = 168 - 70x
y = (-70/21)X + 8  <--- writes the equation of this line in slope intercept form

Fat: 8x + 7y = 24
7y = 24 - 8x
y = (-8/7)X + 24/7   <--- writes the equation of this line in slope intercept form 


Mineral Ash: 7x + 1y = 14
y = - 7x + 14    <--- writes the equation of this line in slope intercept form

The cost function is C(x,y) = $21X + $7Y

Now we need to find where these three lines meet.

A quick formula for doing that is
            x-intersect = (B2-B1)/(M1-M2)
where M1 and M2 are the slopes of the lines and
B1 and B2 are the intercepts of the lines. 

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1st and 2nd lines meet at : (24/7 - 8)/(-70/21 - (-8/7)) = 48/23

Plugging into the first line: (-70/21)(48/23)+8 = 24/23

Plugging into the second line: (-8/7)(48/23) + 24/7 = 24/23

So indeed, the first two lines intersect at (x=48/23, y = 24/23)

The cost for these number of bags is 21(48/23) + 7(24/23) = $51.13
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1st and 3rd lines meet at : (14 - 8) / ((-70/21) - (-7)) = 126/77 

Plugging into the first line: (-70/21)(126/77) + 8 = 588/231 = 2 and 6/11

Plugging into the third line:  -7(126/77) + 14 = 28/11 = 2 and 6/11

So indeed , the first and third lines meet at ( x=126/77, y = 28/11)

The cost for these number of bags is 21(126/77) + 7(28/11) = $52.18

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2nd and 3rd lines meet at : (14 - 24/7) / ( - 8/7 - (-7)) = 74/41

Plugging into the second line: (-8/7)(74/41) + 24/7 = 392/287 = 1 and 15/41

Plugging into the third line: -7(74/41) + 14 = 56/41 = 1 and 15/41

So indeed, the second and third lines meet at ( x = 74/41, y = 56/41)

The cost for these number of bags is 21(74/41) + 7(56/41) = $47.46 

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FINAL ANSWER:
THE WINNER IS OPTION #3.
The third scenario is cheaper.
You need

( x = 74/41, y = 56/41)

x= 74/41 =  1 and 33/41 bags of soybeans (or 1.0848 bags)
y = 56/41 = 1 and 15/41 bags of oats (1.36585 36585 36585 36585..... bags)
I have no idea how you are going to do that.
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