Mrs. Lewis needs to buy two types of grain, oats, and barley to mix as a feed supplement for her cattle. She has $4275 to spend on grain and wants the mixture to be 3 parts oats and 2 parts barley. She can buy oats for $1.10 per bushel and barley for $2.10 per bushel. Mrs. Lewis needs 7,000 bushels of grain. How many bushels of barley should she buy?
Mrs. Lewis cannot buy 7,000 bushels of grain, for that would cost her at least 7,000*$1.10=$7,700, but she only has $4275.
Let X=bushels of oat and Y=bushels of barley. 3 parts oats and 2 parts barley means 2X=3Y, or X=1.5Y. Her cost is 4275=1.1X+2.1Y. Now substitute X=1.5Y and simplify: 4275=1.1(1.5Y)+2.1Y=3.75Y. Therefore, Y=4275/3.75=1140, so she should buy 1140 bushels of barley.
This problem is not a math problem, but an economic's problem. What do you do when you're over budget? There isn't a simple answer to that!
7000 bushels of grain, even if all the grain is the less expensive $1.10/bushel of oats will cost a lot more then $4275.00!
It's time to sell some livestock.
O = # of bushels of oats
B = # of bushels of barley
O+B = 7,000 ......(1)
O = (3/5)*7,000 = 4,200 bushels of oats
B = (2/5)*7,000 = 2,800 bushes of barley
But these cost 4,200*1.10 + 2,800*2.10 = $10,500. So, she needs to borrow 10,500-4,275 = $6,225 to buy 7,000 bushes of grain.
Let's assume, that Mrs. Lewis will buy "x" bushels of oats, and "y" bushes of barley, then:
x + y = 7000 .......... (1)
1.1x + 2.1y = 4275 ....... (2)
xy < 0
(x, y) e Ø
Let x be the oats and y be the barley. Keep the mix requirements with the grain and the $$$ with the $$$.
She needs a total of 7000 bushels of grain which is composed of 3 parts oats and 2 parts barley, thus:
3x + 2y = 7,000
In addition you are told that she has a finite dollar amount she can spend, $4275 where the oats are $1.10/bushel & $2.10/bushel, thus:
1.1x + 2.1y
Now you have 2 unknown values and equations:
3x+2y=7000 (Eq 1)
There are a couple of ways to go about this...find a common coefficient and subtract one equation or solve either equation for x or y and plug it into the other equation. Either way there will be fractions. I am going to solve for y in Eq 1 since it has whole numbers and plug that into Eq 2:
3x+2y=7000 -> 2y=7000-3x -> y= (7000-3x)2 -> y=3500-1.5x (**)
Now plug y into Eq 2
1.1x+2.1y=4275 -> 1.1x + 2.1(3500-1.5x) -> 1.1x + 7350 - 3.15x = 4275 -> 7350 -2.05x = 4275
-> 2.05x = 3075 -> x = 3075/2.05 -> x=1500
Now that you have x=1500 plug that into Eq 1 or Eq 2 to find...you already solved Eq 1 for y earlier (**) so that would be easiest to use:
(**) y = 3500 - 1.5x -> y = 3500 - 1.5(1500) -> y = 3500 - 2250 -> y = 1250
1500 bushels of oats and 1250 bushels of barley. Plug these numbers into either equation to verify. I don't have a calculator on me (hence my taking the whole number path!) so I will leave this to you.
Good luck and I hope it helped!