A farmer is going to divide her 30 acre farm between two crops. Seed for crop A costs $5 per acre. Seed for crop B costs $10 per acre. The farmer can spend at most $200 on seed. If crop B brings in a profit of $50 per acre, and crop A brings in a profit of $160 per acre, how many acres of each crop should the farmer plant to maximize her profit

30 acres of A

0 acres of B

Step-by-step explanation:

On forming the equations we get

Acres = A+B 30  .............(1)

Seed = 5A + 10B 200 ...........(2)

Profit = 160A+50B

Upon rearranging this we get

A= 30-B     sing equation (1)

A=40-2B   using equation (2)

 On plotting these inequalities on the cartesian planes we will get the following points of intersection (0,20) (20,30) (30,0)

Now using these coordinates (A,B) to get Maximum profit

P= $1000 for (0,20)

P=$4700 for (20,30)

P= $4800 for (30,0)

clearly for (30,0) we will get the maximum profit on the crops

step-by-step explanation:

why? because i noticed a pattern. 60 120. those numbers without the zero, make the 6 times tables leaving your answer to be b since 6 times 4 is 24 and adding a zero making it 240.