Introduction to Optimization: Practice

By the corner-point principle, evaluate $P = x + 2y$ at every corner.

$P(6, 0) = 6, \qquad P(4, 3) = 10, \qquad P(0, 4) = 8$

The largest value is $10$, at $(4, 3)$.

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

$(2x + y) - (x + y) = 8 - 5 \;\Rightarrow\; x = 3$

Then $y = 5 - 3 = 2$, so the lines meet at $(3, 2)$.

The total acreage $x + y$ is at most $12$, and the fertilizer used $2x + y$ is at most $18$.

$x + y \le 12, \qquad 2x + y \le 18$

Both resources are upper limits, so both are $\le$ inequalities.

Evaluate $P = 2x + 3y$ at each corner.

$P(4, 0) = 8, \qquad P(3, 2) = 12, \qquad P(0, 3) = 9$

The maximum is $12$, at the inside corner $(3, 2)$.

The maximum sits at the corner $(3, 2)$, so evaluate there.

$P(3, 2) = 2(3) + 3(2) = 6 + 6 = 12$

So the maximum value is $12$.

Subtract the second equation from the first to remove $2y$.

$(3x + 2y) - (x + 2y) = 12 - 8 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = 2$

Then $2 + 2y = 8$ gives $y = 3$, so the meeting point is $(2, 3)$.

Evaluate $P = 2x + y$ at the low corners, where the objective is smallest.

$P(1, 0) = 2, \qquad P(0, 1) = 1, \qquad P(0, 3) = 3$

The smallest value is $1$, at $(0, 1)$.

Substitute the optimal corner into the objective.

$C = 3(4) + 5(1) = 12 + 5 = 17$

So the minimum value is $17$.

On the $x$-axis, $y = 0$. Substitute and solve.

$3x = 12 \;\Rightarrow\; x = 4$

So the corner is $(4, 0)$.

Evaluate $P = 3x + 4y$ at each corner.

$P(8, 0) = 24, \qquad P(6, 4) = 34, \qquad P(0, 7) = 28$

The maximum is $34$, at $(6, 4)$.

Cloth used is $3$ meters per shirt and $2$ meters per pair of pants, at most $24$.

$3x + 2y \le 24$

The coefficients match each item's cloth cost.

Add the two equations to remove $y$.

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

Then $4 - y = 2$ gives $y = 2$, so they meet at $(4, 2)$.