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A farmer is going to divide her $50$ acre farm between two crops. Seed for crop A costs $\$ 50$ per acre. Seed for crop B costs $\$ 25$ per acre. The farmer can spend at most $\$ 2,250$ on seed. If crop B brings in a profit of $\$ 90$ per acre, and crop A brings in a profit of $\$ 220$ per acre, how many acres of each crop should the farmer plant to maximize her profit? Write the objective function then use the feasible region shown in the graph below to maximize it. $\newline$ Let $x=$ the number of acres of crop $A$ $\newline$ Let $y=$ the number of acres of crop $B$ $\newline$ Objective Function: $P=$ $\newline$ Constraints: $\newline$ $\begin{array}{l} x \geq 0, y \geq 0 \\ x+y \leq 50 \\ 50 x+25 y \leq 2250 \end{array}$ $\newline$ The Farmer should Plant $\$ 25$ $0$ acres of crop A and $\$ 25$ $0$ acres of crop B.

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Math Problems

Grade 6

One-step inequalities: word problems

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Q. A farmer is going to divide her

50

acre farm between two crops. Seed for crop A costs

\$ 50

per acre. Seed for crop B costs

\$ 25

per acre. The farmer can spend at most

\$ 2,250

on seed. If crop B brings in a profit of

\$ 90

per acre, and crop A brings in a profit of

\$ 220

per acre, how many acres of each crop should the farmer plant to maximize her profit? Write the objective function then use the feasible region shown in the graph below to maximize it.

\newline

Let

x=

the number of acres of crop

A

\newline

Let

y=

the number of acres of crop

B

\newline

Objective Function:

P=

\newline

Constraints:

\newline

\begin{array}{l} x \geq 0, y \geq 0 \\ x+y \leq 50 \\ 50 x+25 y \leq 2250 \end{array}

\newline

The Farmer should Plant

\$ 25

0

acres of crop A and

\$ 25

0

acres of crop B.

Define variables: Define variables: $\newline$ Let $x$ = number of acres of crop $A$ . $\newline$ Let $y$ = number of acres of crop $B$ .
Write objective function: Write the objective function: $\newline$ Objective Function: $P = 220x + 90y$ $\newline$ This function represents the total profit from $x$ acres of crop A and $y$ acres of crop B.
Write constraints: Write the constraints: $\newline$ $1$ . $x \geq 0$ (Non-negativity constraint for crop A) $\newline$ $2$ . $y \geq 0$ (Non-negativity constraint for crop B) $\newline$ $3$ . $x + y \leq 50$ (Total acres cannot exceed $50$ ) $\newline$ $4$ . $50x + 25y \leq 2250$ (Total cost of seeds cannot exceed $\$2250$ \))
Solve inequalities: Solve the system of inequalities to find feasible region: $\newline$ Using the constraints, plot or calculate the intersection points: $\newline$ - From $x + y = 50$ and $50x + 25y = 2250$ , substitute $y = 50 - x$ into the second equation: $\newline$ $50x + 25(50 - x) = 2250$ $\newline$ $50x + 1250 - 25x = 2250$ $\newline$ $25x = 1000$ $\newline$ $x = 40$ $\newline$ - Substitute $x = 40$ into $y = 50 - x$ : $\newline$ $y = 50 - 40$ $\newline$ $50x + 25y = 2250$ $0$
Check profit maximization: Check if this point $(x=40, y=10)$ maximizes the profit: $\newline$ Substitute $x = 40$ and $y = 10$ into the objective function: $\newline$ $P = 220(40) + 90(10)$ $\newline$ $P = 8800 + 900$ $\newline$ $P = 9700$ $\newline$ This is the maximum profit since other vertices of the feasible region will yield lower profits.