Introduction to Optimization: Additional Practice (Set 2)

The profit is $P = 5x + 4y$, largest at the corner $(3, 3)$.

$P(3, 3) = 5(3) + 4(3) = 15 + 12 = 27$

So the maximum profit is $27$ dollars.

The profit $P = 20x + 15y$ is largest at $(3, 2)$.

$P(3, 2) = 20(3) + 15(2) = 60 + 30 = 90$

So the maximum profit is $90$ dollars.

Check the constraints: $x + y = 6 \le 6$ and $x = 4 \le 4$, so $(4, 2)$ is feasible. Now evaluate the objective.

$P(4, 2) = 3(4) + 2(2) = 12 + 4 = 16$

So the plan is feasible with value $16$.

The region runs up and to the right without bound, so $P = 2x + 3y$ can be made as large as you like: no maximum. Its smallest value sits at the corner $(5, 0)$, since $(0, 5)$ gives $15$ but $(5, 0)$ gives $10$.

$\min P = 10 \text{ at } (5, 0), \qquad \text{no maximum}$

So a minimum exists but no maximum.

The corners are $(0, 0)$, $(6, 0)$, $(0, 4)$, and the intersection of $x + y = 6$ with $y = 4$, which is $(2, 4)$. Evaluate the objective.

$P(6, 0) = 6, \qquad P(2, 4) = 14, \qquad P(0, 4) = 12$

The maximum is at $(2, 4)$.

The corners are $(0, 0)$, $(5, 0)$, $(3, 4)$ (where $2x + y = 10$ meets $x + y = 7$), and $(0, 7)$. Evaluate the objective.

$P(5, 0) = 15, \qquad P(3, 4) = 29, \qquad P(0, 7) = 35$

The maximum value is $35$, at $(0, 7)$.

Both lines have the same slope (rewrite as $y = -x + 4$ and $y = -x + 7$) but different intercepts.

$x + y = 4 \; \parallel \; x + y = 7$

Parallel lines never cross, so there is no intersection point.

If the level lines match the slope of an edge, the last one to touch the region lies flat along that whole edge.

$\text{objective parallel to an edge} \;\Rightarrow\; \text{every point of that edge is optimal}$

Then both corners at the ends of the edge, and every point between them, tie for the best value.

Test each point in $2x + 3y \le 12$.

$2(2) + 3(3) = 4 + 9 = 13 > 12$

The point $(2, 3)$ breaks the constraint. The others give $12$, $12$, and $12$, so they are all feasible.

Evaluate $P = x + 4y$ at the corners $(5, 0)$, $(5, 1)$, $(3, 3)$, and $(0, 3)$.

$P(5, 1) = 9, \qquad P(3, 3) = 15, \qquad P(0, 3) = 12$

The maximum is $15$, at $(3, 3)$. Changing the objective moved the best corner away from $(5, 1)$.

The total investment is at most $10$, so $x + y \le 10$, and "at least $2$ in bonds" means $y \ge 2$.

$x + y \le 10, \qquad y \ge 2$

The total is a $\le$ limit (not forced equal), and the bond minimum is a $\ge$ limit.

The corner-point principle guarantees the maximum sits at a vertex, so checking every vertex cannot miss it.

$\text{evaluate } P \text{ at every vertex, then take the largest}$

Testing the center, random interior points, or the corner nearest the origin can all miss the true best corner.