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?

savcarroll Apr 25, 2019

#1**+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.**

CalculatorUser Apr 25, 2019

#2**+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.**

Guest Apr 25, 2019