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A farmer has 1,300 acres of land on which he grows corn, wheat, and soybeans. It costs $45 per acre to grow corn, $60 to grow wheat, and $50 to grow soybeans. Because of market demand the farmer will grow twice as many acres of wheat as of corn. He has allocated $69,200 for the cost of growing his crops. How many acres of each crop should he plant?

 Apr 25, 2019
 #1
avatar+2547 
+5

This is an algebra problem that can be solved by making an equation with 3 variables.

 

(1) Make Equations

 

First equation: "It costs $45 per acre to grow corn, $60 to grow wheat and $50 to grow soybeans" = "A farmer has 1,300 acres of land on which he grows corn, wheat, and soybeans."

 

So the first equation is this  : \(x+y+z=1300\) the three variables representing HOW MANY acres

 

Second equation : "It costs $45 per acre to grow corn, $60 to grow wheat and $50 to grow soybeans" = "He has allocated $69,200 for the cost of growing his crops"

 

So the second equation is this : \(45x+60y+50z=69200\)

 

Third equation : "Because of market demand the farmer will grow twice as many acres of wheat as of corn."

 

So the third equation is this : \(y=2x\)

 

(2) Solve

 

Now we know the two equations, we can solve.

 

We can substitute the third equation into the first equation to get \(3x+z=1300\)

 

Now isolate z.

 

\(z=1300-3x\)

 

We can also substitute the third equation into the second equation.

 

\(165x+50z=69200\)

 

Now substitute the equation \(z=1300-3x\) into the equation above.

 

After all this, you should get the answer of 

 

\(\boxed{x=280}\)

 

\(\boxed{y=560}\)

 

\(\boxed{z=460}\)

 

So in the end, we should have 280 acres of corn, 560 acres of wheat, and 460 acres of soybeans.

 Apr 25, 2019
 #3
avatar+106539 
+1

Very nice explanation, CU  !!!!

 

I'm impressed with the clarity of your answer.......

 

 

cool cool cool

CPhill  Apr 25, 2019
 #2
avatar
+2

Let the acreage reseved for corn =C
The acreage reserved for wheat   =2C
The acreage reserved for soybeans=1,300 - C - 2C
45C + 120C + 50[1,300 - C - 2C] = $69, 200, solve for C
C = 280 - acres reserved for corn
280 x 2 =560 acres reserved for wheat
1300 - 280 - 560 =460 acres reserved for soybeans.

 Apr 25, 2019

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