a farmer has 1200 acres of land and plans to plant corn and soybeans. the input cost (cost of seed…

a farmer has 1200 acres of land and plans to plant corn and soybeans. the input cost (cost of seed, fertilizer, herbicide, and insecticide) for 1 acre for each crop is given in the table along with the cost of machinery and labor. the profit for 1 acre of each crop is given in the last column. suppose the farmer has budgeted a maximum of $172,000 for input costs and a maximum of $104,000 for labor and machinery. \n| | corn | soybeans |\n|--|--|--|\n| input cost per acre | $160 | $120 |\n| labor/machinery cost per acre | $60 | $100 |\n| profit per acre | $120 | $100 |\npart: 0 / 3\npart 1 of 3\n(a) determine the number of acres of each crop that the farmer should plant to maximize profit. (assume that all crops will be sold.)\nthe farmer should plant acres of corn and acres of soybeans.

a farmer has 1200 acres of land and plans to plant corn and soybeans. the input cost (cost of seed, fertilizer, herbicide, and insecticide) for 1 acre for each crop is given in the table along with the cost of machinery and labor. the profit for 1 acre of each crop is given in the last column. suppose the farmer has budgeted a maximum of $172,000 for input costs and a maximum of $104,000 for labor and machinery. \n| | corn | soybeans |\n|--|--|--|\n| input cost per acre | $160 | $120 |\n| labor/machinery cost per acre | $60 | $100 |\n| profit per acre | $120 | $100 |\npart: 0 / 3\npart 1 of 3\n(a) determine the number of acres of each crop that the farmer should plant to maximize profit. (assume that all crops will be sold.)\nthe farmer should plant acres of corn and acres of soybeans.

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of acres of corn and $y$ be the number of acres of soybeans.

Step2: Set up constraints

We have the following constraints:

  • Land - constraint: $x + y\leq1200$
  • Input - cost constraint: $160x + 120y\leq172000$
  • Labor - machinery cost constraint: $60x+100y\leq104000$
  • Non - negativity constraints: $x\geq0,y\geq0$

Step3: Rewrite the inequalities

The input - cost inequality $160x + 120y\leq172000$ simplifies to $4x + 3y\leq4300$. The labor - machinery cost inequality $60x + 100y\leq104000$ simplifies to $3x+5y\leq5200$.

Step4: The objective function

The profit function $P$ that we want to maximize is $P = 120x+100y$.

Step5: Solve the system of inequalities

First, consider the intersection of $x + y=1200$ and $4x + 3y=4300$. From $x + y=1200$, we have $y = 1200 - x$. Substitute into $4x + 3y=4300$: $4x+3(1200 - x)=4300$ $4x + 3600-3x=4300$ $x=700$, then $y = 1200 - 700=500$. Next, consider the intersection of $x + y=1200$ and $3x + 5y=5200$. From $x + y=1200$, $x=1200 - y$. Substitute into $3x + 5y=5200$: $3(1200 - y)+5y=5200$ $3600-3y + 5y=5200$ $2y=1600$ $y = 800$, then $x=1200 - 800 = 400$. Finally, consider the intersection of $4x + 3y=4300$ and $3x + 5y=5200$. Multiply the first equation by 3 and the second by 4: $12x+9y = 12900$ and $12x+20y=20800$. Subtract the first from the second: $(12x + 20y)-(12x + 9y)=20800 - 12900$ $11y=7900$ $y=\frac{7900}{11}\approx718.18$, then $x=\frac{4300-3\times\frac{7900}{11}}{4}=\frac{4300\times11-23700}{44}=\frac{47300 - 23700}{44}=\frac{23600}{44}\approx536.36$. Evaluate the profit function $P = 120x+100y$ at the corner - points:

  • At $(0,0)$, $P=0$.
  • At $(0, 1040)$ (from $3x + 5y=5200$ when $x = 0$), $P=100\times1040 = 104000$.
  • At $(1075,0)$ (from $4x + 3y=4300$ when $y = 0$), $P=120\times1075=129000$.
  • At $(400,800)$: $P=120\times400+100\times800=48000 + 80000=128000$.
  • At $(700,500)$: $P=120\times700+100\times500=84000+50000 = 134000$.

Answer:

700 acres of corn and 500 acres of soybeans.