There are two decision variables:Originally Posted byLilDragonfly

hectares of apples, A

hectares of pears, P

The constraints are:

A +P <= 10 ----------------------(1)

5000A +20,000P <= 100,000

Divide both sides by 5000,

A +4P <= 20 ----------------------(2)

Non-negative, non-zero constraints:

A > 0 ------------------------------(3)

P > 0 ---------------------------(4)

The objective function is

Profit = 4000A +8000P -----------***

Now, plot the inequalities (1), (2), (3) and (4) on the same (A,P) cartesian or rectangular coordinate axes.

Get the intersection points of the 4 inequalities. Since A or P cannot be zero, then the intersection of (1) and (2) , being the only intersection point left, should determine the optimum (maximum here) profit.

The intersection of (1) and (2):

Consider them as equations only.

(2) minus (1),

3P = 10

So, P = 10/3

And, using (1), A = 10 -P = 10 -(10/3) = 20/3

Therefore, the intersection of (1) and (2), and at the same time the only effective corner of the feasible/solution region, is (20/3, 10/3).

Plugging that into the objective function,

Profit = 4000A +8000P

Maximum profit = 4000(20/3) +8000(10/3) = 160,000/3 = $53,333.33 ----------answer.

To achieve that, the orchardist should plant (6 and 2/3) hectares of apples and (3 and 1/3) hectares of pears. ------------------answer.