Introduction to Optimization: Practice

Substitute $x = 3$ and $y = 0$.

$P = 2(3) + 5(0) = 6 + 0 = 6$

Because $y = 0$, only the $x$ term contributes.

Work out the objective at each point.

$C(2, 1) = 2 + 4 = 6, \qquad C(5, 0) = 5 + 0 = 5$

Since $5 < 6$, the point $(5, 0)$ gives the smaller value.

Add the coordinates at each point.

$P(3, 2) = 5, \qquad P(1, 5) = 6$

Since $6 > 5$, the objective is larger at $(1, 5)$.

A linear objective is pushed to its extreme by sliding its parallel level lines, and the last one to touch the region meets it at a corner.

$\text{the maximum and the minimum occur at a vertex}$

So you only need to check the corners, not the interior.

The flour used is $2$ cups for each of the $x$ cakes and $1$ cup for each of the $y$ muffins, and it cannot exceed $10$.

$2x + y \le 10$

The "at most" makes it a $\le$ inequality.

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

$P(4, 0) = 8, \qquad P(3, 3) = 9, \qquad P(0, 4) = 4$

The largest value is $9$, at $(3, 3)$.

Evaluate $P = 3x + y$ at the labeled corners.

$P(5, 0) = 15, \qquad P(3, 3) = 12, \qquad P(0, 2) = 2$

The heavy weight on $x$ makes the corner $(5, 0)$ win with $15$.

The point must have $x = 0$ (on the $y$-axis) and $y = 5$ at the same time.

$x = 0, \; y = 5 \;\Rightarrow\; (0, 5)$

So the lines cross at $(0, 5)$.

Substitute $x = 1$ and $y = 2$.

$C = 4(1) + 3(2) = 4 + 6 = 10$

So the objective is worth $10$ there.

Evaluate $C = 2x + 2y$ at each corner.

$C(1, 4) = 10, \qquad C(3, 1) = 8, \qquad C(5, 2) = 14$

The smallest value is $8$, at $(3, 1)$.

Evaluate the objective at each point.

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

The large coefficient on $x$ makes $(2, 0)$ the larger of the two.

Each bracelet adds $3$ and each necklace adds $7$ to the earnings.

$P = 3x + 7y$

This linear expression is the objective to maximize.