a farmer has 1400 acres of land and plans to plant corn and soybeans. the input cost (cost of seed…

a farmer has 1400 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 $216,000 for input costs and a maximum of $131,000 for labor and machinery. part: 0 / 3 part 1 of 3 (a) determine the number of acres of each crop that the farmer should plant to maximize profit. (assume that all crops will be sold.) the farmer should plant acres of corn and acres of soybeans.

a farmer has 1400 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 $216,000 for input costs and a maximum of $131,000 for labor and machinery. part: 0 / 3 part 1 of 3 (a) determine the number of acres of each crop that the farmer should plant to maximize profit. (assume that all crops will be sold.) the 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: $x + y\leq1400$
  • Input - cost constraint: $180x + 120y\leq216000$ (divide by 60 gives $3x + 2y\leq3600$)
  • Labor - machinery cost constraint: $70x+100y\leq131000$ (divide by 10 gives $7x + 10y\leq13100$)
  • And $x\geq0,y\geq0$

Step3: Set up the objective function

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

Step4: Solve the system of inequalities

First, rewrite the inequalities as equations to find the corner - points:

  • $x + y=1400$ gives $y = 1400 - x$
  • $3x + 2y=3600$, substitute $y=1400 - x$: $3x+2(1400 - x)=3600$, $3x + 2800-2x=3600$, $x = 800$, then $y=1400 - 800 = 600$
  • $7x + 10y=13100$, substitute $y = 1400 - x$: $7x+10(1400 - x)=13100$, $7x+14000-10x=13100$, $- 3x=-900$, $x = 300$, then $y=1400 - 300 = 1100$ Check the intersection of $3x + 2y=3600$ and $7x + 10y=13100$. Multiply $3x + 2y=3600$ by 5 to get $15x+10y = 18000$. Subtract $7x + 10y=13100$ from it: $(15x + 10y)-(7x + 10y)=18000 - 13100$, $8x=4900$, $x=\frac{4900}{8}=612.5$, then $y=\frac{3600-3\times612.5}{2}=\frac{3600 - 1837.5}{2}=881.25$ The corner - points of the feasible region are $(0,0),(0,1100),(300,1100),(800,600),(1200,0)$

Step5: Evaluate the objective function at corner - points

  • $P(0,0)=120\times0 + 100\times0=0$
  • $P(0,1100)=120\times0+100\times1100 = 110000$
  • $P(300,1100)=120\times300+100\times1100=36000 + 110000=146000$
  • $P(800,600)=120\times800+100\times600=96000+60000 = 156000$
  • $P(1200,0)=120\times1200+100\times0=144000$

Answer:

The farmer should plant 800 acres of corn and 600 acres of soybeans.