it can be solved graphically as long as there are only two unknown variables involved.
need an objective function and some constraint functions.
let x = the maximum number of tons of fertilizer from plant A each week.
let y = the maximum number of tons of fertilizer from plant B each week.
the manager wants to make a total of at least 70 tons of fertilizer from both plants each week.
your constraints are:
x <= 40
y <= 50
x + y >= 70
x >= 0
y >= 0
your objective function will be the number of pounds of particulate matter placed into the atmosphere each week.
plant A places 20 pounds of particulate matter per ton of fertilizer produced.
plant B places 40 pounds of particulate matter per ton of fertilizer produced.
your objective function is y = 20x + 40y
that is the function that you want to minimize.
using the desmos.com calculator, you would graph the opposite of the constraint functions.
the are of the graph that is not shaded will be the feasible region.
you would then evaluate the objective function at each of the corner points of the feasible region to find the minimum amount of particulate placed into the atmosphere.
my graph looks like this:
the evaluation of at the corner points is shown in the following table.
(x,y) 20x + 40y (20,50) 2400 (40,50) 2800 (40,30) 2000
the minimum pounds of particulate placed into the atmosphere is when plant A produces 40 tons of fertilizer and when plant B produces 30 tons of fertilizer each week.
the constraints for the minimum particulate solution have to be satisfied as well.
x <= 40 is satisfied because x = 40.
y <= 50 is satisfied because y = 30
x + y >= 70 is satisfied because x + y = 70
x >= 0 and y >= 0 are both satisfied because x and y are both > 0.
the desmos.com calculator can be found at https://www.desmos.com/calculator