+24 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?
+2863 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.
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.
