Introduction to Optimization: Additional Practice (Set 1)

The corners are $(0, 0)$, $(7, 0)$, $(0, 5)$, and the intersection of $x + y = 7$ with $x + 3y = 15$, which is $(3, 4)$. Evaluate the objective.

$P(7, 0) = 35, \qquad P(3, 4) = 27, \qquad P(0, 5) = 15$

The maximum is at $(7, 0)$; the heavy weight on $x$ makes the far $x$-axis corner win.

Evaluate $C = x + 6y$ at each corner.

$C(0, 5) = 30, \quad C(1, 2) = 13, \quad C(4, 1) = 10, \quad C(6, 0) = 6$

The smallest value is $6$, at $(6, 0)$.

Subtract the second equation from the first to remove $y$.

$(4x + y) - (x + y) = 14 - 8 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$

Then $2 + y = 8$ gives $y = 6$, so the meeting point is $(2, 6)$.

Using all the hours means $x + y = 5$ (wiring) and $2x + y = 8$ (finishing). Subtract the first from the second.

$x = 3, \quad y = 5 - 3 = 2$

So the corner where both limits bind is $(3, 2)$.

Test each objective at the three outer corners; you want the one that peaks at $(6, 0)$.

$3x + y: \; (6, 0) \to 18, \quad (4, 4) \to 16, \quad (0, 6) \to 6$

Only $3x + y$ makes $(6, 0)$ the winner. The others peak at $(4, 4)$ or $(0, 6)$.

Test each point in both inequalities.

$2(2) + 2 = 6 \le 6, \qquad 2 + 2(2) = 6 \le 6$

So $(2, 2)$ works. The point $(3, 1)$ fails the first, $(1, 3)$ fails the second, and $(0, 4)$ fails the second since $8 > 6$.

Evaluate $C = 4x + 5y$ at each corner.

$C(0, 10) = 50, \qquad C(2, 4) = 28, \qquad C(6, 0) = 24$

The smallest value is $24$, at $(6, 0)$.

Subtract the first equation from the second to remove $3y$.

$(4x + 3y) - (2x + 3y) = 18 - 12 \;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3$

Then $2(3) + 3y = 12$ gives $3y = 6$, so $y = 2$ and the point is $(3, 2)$.

Read the corners from the graph and evaluate $P = x + 4y$ at each.

$P(5, 0) = 5, \qquad P(3, 3) = 15, \qquad P(0, 2) = 8$

The largest value is $15$, at $(3, 3)$.

Read the corners from the graph and evaluate $P = x + 2y$ at each.

$P(2, 0) = 2, \quad P(5, 1) = 7, \quad P(3, 4) = 11, \quad P(0, 2) = 4$

The largest value is $11$, at $(3, 4)$.

The constraints are $2x + y \le 16$ and $x + y \le 10$, giving corners $(0, 0)$, $(8, 0)$, $(6, 4)$, and $(0, 10)$. Evaluate the profit $P = 6x + 5y$.

$P(8, 0) = 48, \qquad P(6, 4) = 56, \qquad P(0, 10) = 50$

The maximum is at $(6, 4)$, where both resources are fully used.

The corners are $(0, 0)$, $(5, 0)$, $(5, 1)$ (where $x = 5$ meets $x + y = 6$), $(3, 3)$ (where $y = 3$ meets $x + y = 6$), and $(0, 3)$. Evaluate the objective.

$P(5, 0) = 20, \qquad P(5, 1) = 21, \qquad P(3, 3) = 15$

The maximum is at $(5, 1)$.