A farmer has 70 acres of land available for planting either...

A farmer has 70 acres of land available for planting either soybeans or wheat. The cost of preparing the soil, the workdays required, and the expected profit per acre planted for each type of crop are given in the following table. $$ \begin{array}{|lcc|} \hline & \text { Soybeans } & \text { Wheat } \\ \hline \text { Preparation cost per acre } & \$ 60 & \$ 30 \\ \text { Workdays required per acre } & 3 & 4 \\ \text { Profit per acre } & \$ 180 & \$ 100 \\ \hline \end{array} $$ The farmer cannot spend more than $\$ 1800$ in preparation costs and cannot use a total of more than 120 workdays. How many acres of each crop should be planted to maximize the profit? What is the maximum profit? What is the maximum profit if the farmer is willing to spend no more than $\$ 2400$ on preparation?

A farmer has 70 acres of land available for planting either soybeans or wheat. The cost of preparing the soil, the workdays required, and the expected profit per acre planted for each type of crop are given in the following table.
$$
\begin{array}{|lcc|}
\hline & \text { Soybeans } & \text { Wheat } \\
\hline \text { Preparation cost per acre } & \$ 60 & \$ 30 \\
\text { Workdays required per acre } & 3 & 4 \\
\text { Profit per acre } & \$ 180 & \$ 100 \\
\hline
\end{array}
$$
The farmer cannot spend more than $\$ 1800$ in preparation costs and cannot use a total of more than 120 workdays. How many acres of each crop should be planted to maximize the profit? What is the maximum profit? What is the maximum profit if the farmer is willing to spend no more than $\$ 2400$ on preparation?

Key Concepts

Linear Programming

Linear programming is a mathematical method used to determine the best possible outcome in a given mathematical model whose requirements are represented by linear relationships. It involves optimizing a linear objective function, subject to a set of linear inequality or equality constraints that represent limitations on resources or conditions that must be met.

Decision Variables

Decision variables are the quantities that the decision maker will decide the values of in order to achieve the best outcome in a linear programming problem. They represent the choices available, and the goal is to determine the optimal values of these variables that maximize or minimize the objective function.

Objective Function

The objective function in a linear programming problem is the function that needs to be optimized, either maximized or minimized. It is a linear function of the decision variables and represents the goal of the problem, such as maximizing profit or minimizing cost.

Constraints

Constraints are the restrictions or limitations on the decision variables in a linear programming problem. They are expressed as linear inequalities or equalities and define the feasible region within which a solution must lie by representing limits on resources, capacities, or other factors.

Feasible Region

The feasible region is the set of all possible points that satisfy all the constraints of a linear programming problem. It is typically visualized in a coordinate system and the optimal solution is found at one of the vertices or along a boundary segment of this region.

Graphical Method

The graphical method is a technique used to solve linear programming problems with two decision variables by plotting the constraints on a graph. The feasible region is identified visually, and the optimal solution is determined by evaluating the objective function at the corner points of this region.

Sensitivity Analysis

Sensitivity analysis involves studying how the changes in a linear programming model—such as changes in constraint limits or coefficients in the objective function—affect the optimal solution. It provides insight into the robustness and reliability of the chosen solution under varying conditions or resource availability.

Transcript

00:01 So if the farmer at most can spend $1 ,800 and can spend 120 work days, what is the maximum profit that can be yielded given that information? so we're going to begin by labeling some variables.

00:19 So x will call number of acres of soybeans.

00:27 Likewise, we'll call y the number of acres of wheat planted.

00:35 And z will be our maximum profit.

00:42 And now we can write our objective function, which would start with z, our profit, and since we have $180 profit for x, we do 180x plus 100y.

00:58 So here's our objective function.

01:01 And now what we want to do is write down all of our constraints.

01:05 So we know that x and y must be non -negative, so x, is greater than or equal to 0, and y is also greater than or equal to 0.

01:15 And we're given our limits up here.

01:18 So we know that 60x, which is the cost of preparing the soybeans, plus 30y, which is the cost of preparing the wheat, cannot be more than, so must be less than or equal to 1 ,800, which is the amount of money that the farmer is willing to spend.

01:39 Likewise, we have three workdays for soybeans and four work days for wheat.

01:45 So 3x plus 4 y must be less than or equal to 120.

01:53 So these are our constraints and now we just have to graph them.

01:58 So x and y being greater than are equal to 0 tells us that our graph will be in the first quadrant.

02:03 And i can rearrange this equation to be 30y is less than than or equal to negative 60x plus 1 ,800.

02:14 I divide both sides by 30 and i get y is less than or equal to negative 2x plus 60.

02:23 Now this is much easier to graph because i know my y intercept is at 60 and since i have a slope of negative 2 my x intercept will be at 30.

02:35 Moving on i can rewrite this as 4y is less than or equal to negative 3x plus 120.

02:44 I divide both sides by 4 to isolate the y, and i get y is less than or equal to negative 3 fourths x plus 30.

02:53 Now i can graph this knowing that my y intercept is at 30, and since i have a slope of negative 3 fourths, i know that my x intercept will be at 40.

03:10 So now i can shade in, the graph of my feasible points.

03:17 And what my next step would be is to write down the coordinates of all the corner points.

03:24 So starting at the origin, i'll write down 0 .0.

03:29 Moving here, we would have 0 .30.

03:33 Here we would have 30, 0.

03:37 And to get this point, i'm going to set this equation equal to this equation.

03:43 So we get negative 2x plus 60 equals negative 3 4ths x plus 30.

03:52 And i just want to isolate x and i can start by multiplying both sides by 4.

03:59 To get negative 8x plus 240 equals negative 3x plus 120.

04:07 I can add 5x to both sides and subtract 120 from both sides.

04:12 So x equals 24.

04:18 And now i can plug this into this equation, for example, or this equation, but i think the top one would be easier.

04:27 And to solve for y, i have negative 48 plus 60, so y equals 12...

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