Farmer Jones must determine how many acres of com and wheat to plant this year. An acre of wheat yields 25 bushels of wheat and requires 10 hours of labor per week. An acre of corn yields 10 bushels of corn and requires 4 hours of labor per week. All wheat can be sold at $4 a bushel, and all com can be sold at$3 a bushel. Seven acres of land and 40 hours per week of labor are available. Government regulations requite that at least 30 bushels of corn be produced during the current year. Let x1 = number of acres of corn planted, and x2 = number of acres of wheat planted. Using these decision variables, formulate an LP whose solution will tell Farmer Jones how to maximize the total revenue from wheat and corn.

$\text{\underline{\color{#4257b2}Decision variables}}$

The decision variables are already given.

$\text{\underline{\color{#4257b2}Objective function}}$

The profits need to be maximized. Since $x_1$ acres of wheat yields $25x_1$ bushels of wheat, which sell for $4$ dollars per bushel, so the revenue from $x_1$ acres of wheat is $4 \cdot 25x_1 = 100x_1$ dollars.

Similarly, the revenue from $x_2$ acres of corn is $3 \cdot 10 x_2 = 30 x_2$ dollars.

Therefore, the objective function is

$$\max \ z = 100x_1 + 30x_2$$

$\text{\underline{\color{#4257b2}Constraints}}$

Constraint 1:. Only 7 acres of land are available. Therefore, $x_1 + x_2 \leqslant 7$. Constraint 2:. 40 hours of labor per week are available. Since $x_1$ acres of wheat need $10x_1$ hours of labor per week, and $x_2$ acres of corn need $4x_2$ hours of labor per week, we have $10x_1 + 4x_2 \leqslant 40$. Constraint 3:. The farmer needs to produce at least 30 bushels of corn. Since each acre of corn produces $10$ bushels of corn, we have $10x_2 \geqslant 30$, which can be simplified to $x_2 \geqslant 3$.

$\text{\underline{\color{#4257b2}Sign restrictions}}$

Obviously, $x_1 \geqslant 0$ and $x_2 \geqslant 0$ (acres of corn and wheat must both be nonnegative).

$\text{\underline{\color{#4257b2}Conclusion}}$

Finally, we can write the LP problem as

$$\begin{align*} \max \ z &= 100x_1 + 30x_2 \tag{Objective function}\\ x_1 + \phantom{0}x_2 &\leqslant \phantom{0}7 \tag{Land restriction}\\ 10x_1 + 4x_2 &\leqslant 40 \tag{Labor restriction}\\ x_2 & \geqslant \phantom{0}3 \tag{Corn restriction}\\ x_1 \phantom{+0x_2} & \geqslant \phantom{0}0 \tag{Sign restriction} \\ x_2 & \geqslant \phantom{0}0 \tag{Sign restriction} \\ \end{align*}$$