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A veterinarian needs 60 pounds of dog food that is 15% protein. He will combine a beef mix that is 18% protein with a bacon mix that is 9% protein. How many pounds of each does he need to make the 15% protein mix. Define your variables and write a system of equations to solve.

Answers
Let x be the pounds of beef mix and y the pounds of bacon mix.
advmc
24 January 2012
We know that:
advmc
24 January 2012
x+y=60 (first equation in the system)
advmc
24 January 2012
and that:
advmc
24 January 2012
0.18x+0.9y=0.15*(x+y) (second equation in the system)
advmc
24 January 2012
sorry I mean: 0.18x+0.09y=0.15(x+y)
advmc
24 January 2012
Then , to solve the system, substitute y=60-x in the second equation , to get
advmc
24 January 2012
0.18x+0.09(60-x)=0.1560
advmc
24 January 2012
0.18x+5.4-0.09x=9
advmc
24 January 2012
0.09x=3.6
advmc
24 January 2012
x=40 and therefore y=20
advmc
24 January 2012
This is the mechanical procedure to solve a system of two equations
advmc
24 January 2012
A more intuitive way to solve it: notice that x (beef mix) has twice as much protein as y (bacon mix)
advmc
24 January 2012
rewrite the second equation as:
advmc
24 January 2012
0.09x+0.09(x+y)=0.15(x+y)
advmc
24 January 2012
0.09x=0.06(x+y)
advmc
24 January 2012
then, knowing that x+y=60, you get:
advmc
24 January 2012
0.09x=3.6
advmc
24 January 2012
which leads again to x=40 pounds and therefore y=20 pounds. Notice how x is twice as big as y, just as the proportions between the protein contents
advmc
24 January 2012
Everything clear so far?
advmc
24 January 2012
This is basically a problem of finding the right proportions. It is that relevant to know that we need 60 pounds of dog food, rather, that we need a 15% protein mix.
advmc
24 January 2012
Let z be the proportion of beef mix in the dog food and of course (1-z) is necessarily the proportion of bacon mix. We could have alternatively solved the problem this way, in terms of proportions:
advmc
24 January 2012
z0.18+(1-z)0.09=0.15
advmc
24 January 2012
0.09z=0.06 which means z=2/3 and 1-z=1/3
advmc
24 January 2012
0.09z=0.15-0.09
advmc
24 January 2012
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