Introduction to Optimization: Practice

The optimum sits at $(3, 2)$, so evaluate the objective there.

$P(3, 2) = 4(3) + 3(2) = 12 + 6 = 18$

So the maximum value is $18$.

The two lower-limit constraints leave a region that runs up and to the right forever, so $C = 3x + 2y$ grows without bound and has no maximum. Its smallest value sits at the corner $(2, 2)$, where the two slanted lines cross.

$\min C = 3(2) + 2(2) = 10, \qquad \text{no maximum}$

An unbounded region keeps the minimum here but loses the maximum.

Using all the sugar and all the time means $3x + y = 12$ and $x + y = 6$. Subtract the second from the first.

$2x = 6 \;\Rightarrow\; x = 3, \quad y = 6 - 3 = 3$

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

For $(2, 3)$ to beat both other corners, test each objective there against $(0, 4)$ and $(5, 0)$.

$2x + 3y: \; (0, 4) \to 12, \quad (2, 3) \to 13, \quad (5, 0) \to 10$

Only $2x + 3y$ peaks at $(2, 3)$; $3x + y$ and $4x + y$ peak at $(5, 0)$, and $x + 4y$ peaks at $(0, 4)$.

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

$C(0, 4) = 16, \qquad C(1, 2) = 13, \qquad C(4, 0) = 20$

The smallest value is $13$, at $(1, 2)$.

The corner-point principle guarantees the maximum sits at a vertex, so only the corners need testing.

$\text{a pentagon has } 5 \text{ vertices}$

Evaluate the objective at all $5$ and take the largest.

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

$C(2, 0) = 4, \quad C(0, 2) = 2, \quad C(3, 4) = 10, \quad C(5, 1) = 11$

The smallest value is $2$, at $(0, 2)$.

Profit adds $60$ per jacket and $40$ per pair of trousers, and more profit is better, so it is a maximization.

$\text{maximize } P = 60x + 40y$

The coefficients must match each item's profit.

The optimum is at $(2, 4)$, so evaluate there.

$P(2, 4) = 2 + 3(4) = 2 + 12 = 14$

So the maximum value is $14$.

The profit is $P = 8x + 5y$, largest at $(3, 4)$.

$P(3, 4) = 8(3) + 5(4) = 24 + 20 = 44$

So the maximum profit is $44$ dollars.

At an interior point there is room to move a little in the direction that increases the objective while staying inside the region, and that nearby feasible point has a larger value.

$\text{any interior point is beaten by a nearby feasible point}$

So the maximum is forced onto the boundary, and from there onto a corner.

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

$C(1, 5) = 23, \quad C(2, 2) = 14, \quad C(5, 1) = 19, \quad C(6, 3) = 30$

The smallest value is $14$, at $(2, 2)$.