Answer:
(a) {0 ≤ x ≤ 60; 0 ≤ y ≤ 90; x + y ≤ 120}
(b) p = 50x +75y; 30 tons of loam and 90 tons of peat will maximize profit
Explanation:
You want a graph of constraints and the solution to the problem of maximizing profit from sale of 60 tons of loam and 90 tons of peat at a profit of $50/t and $75/t, respectively, if at most 120 tons can be sold.
(a) Constraints
Let x and y represent tons of loam and peat, respectively. Then the constraints can be written as the inequalities ...
- 0 ≤ x ≤ 60 . . . . . can only sell as much as is available
- 0 ≤ y ≤ 90 . . . . . can only sell as much as is available
- x + y ≤ 120 . . . . . can sell a maximum of 120 tons
(b) Objective function
We want to maximize profit, so maximize ...
p = 50x +75y
Graph
The attached graph shows the graphs of the inequalities. The restrictions on the variables confine the applicable graph to the first quadrant.
The black line shows the objective function. The value of the profit function is maximized when the line is as far from the origin as possible. The boundary point of the constraint solution space that maximizes profit is (x, y) = (30, 90).
Sale of 30 tons of loam and 90 tons of peat soil will maximize profit.
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Additional comment
Since selling peat is the most profitable, profit will be maximized when we sell as much peat as possible. Sale of only 90 tons is possible, so profit will be maximized when we sell that many, then make up the rest of the 120 ton maximum by selling loam. We don't need a graph or a list of constraints to figure that out.