Linear Programming

The feasible region is the triangle with corner points $(0,0)$, $(4,0)$ and $(0,4)$.

Evaluate $Z=3x+4y$ at each corner point:

- At $(0,0)$: $Z=3(0)+4(0)=0$
- At $(4,0)$: $Z=3(4)+4(0)=12$
- At $(0,4)$: $Z=3(0)+4(4)=16$

The maximum value is $16$ at $(0,4)$.

First find the feasible region from the constraints:

- $x+2y\le 8$
- $3x+2y\le 12$
- $x\ge 0,\ y\ge 0$

Corner points are:
- $(0,0)$
- $(0,4)$ from $x=0$ in $x+2y=8$
- $(4,0)$ from $y=0$ in $3x+2y=12$
- Intersection of $x+2y=8$ and $3x+2y=12$:

Subtracting, $2x=4$, so $x=2$.
Then $2+2y=8\Rightarrow y=3$.
So intersection is $(2,3)$.

Now evaluate $Z=-3x+4y$:
- At $(0,0)$: $Z=0$
- At $(0,4)$: $Z=16$
- At $(4,0)$: $Z=-12$
- At $(2,3)$: $Z=-6+12=6$

The minimum value is $-12$ at $(4,0)$.

The constraints are:

- $3x+5y\le 15$
- $5x+2y\le 10$
- $x\ge 0,\ y\ge 0$

Corner points of the feasible region:
- $(0,0)$
- $(0,3)$ from $3x+5y=15$ with $x=0$
- $(2,0)$ from $5x+2y=10$ with $y=0$
- Intersection of $3x+5y=15$ and $5x+2y=10$:

Multiply first by 2: $6x+10y=30$
Multiply second by 5: $25x+10y=50$
Subtract: $19x=20\Rightarrow x=\frac{20}{19}$.

Then $5\cdot\frac{20}{19}+2y=10$ gives $2y=\frac{90}{19}$, so $y=\frac{45}{19}$.

Now compute $Z=5x+3y$:
- $(0,0)$: $0$
- $(0,3)$: $9$
- $(2,0)$: $10$
- $(20/19,45/19)$: $\frac{100}{19}+\frac{135}{19}=\frac{235}{19}$.

Since the Chapter examples and the feasible polygon here give the largest value at the intersection, the maximum is $\frac{235}{19}$.

This exact value is not among the chapter's printed options because no options were printed in the exercise; the computed answer is $\frac{235}{19}$.

We need the minimum of $Z=3x+5y$ subject to

- $x+3y\ge 3$
- $x+y\ge 2$
- $x\ge 0,\ y\ge 0$

Find corner points of the feasible region.

1. Intersection of $x+3y=3$ and $x+y=2$:

Subtract the second from the first:
$2y=1\Rightarrow y=\frac12$.
Then $x+\frac12=2\Rightarrow x=\frac32$.
So one corner point is $(\frac32,\frac12)$.

2. On the $y$-axis ($x=0$):
- From $x+3y\ge 3$, we get $y\ge 1$.
- From $x+y\ge 2$, we get $y\ge 2$.
So the lowest point on the axis is $(0,2)$.

3. On the $x$-axis ($y=0$):
- From $x+3y\ge 3$, we get $x\ge 3$.
- From $x+y\ge 2$, we get $x\ge 2$.
So the lowest point on the axis is $(3,0)$.

Now evaluate $Z$:
- At $(\frac32,\frac12)$: $Z=3\cdot\frac32+5\cdot\frac12=\frac92+\frac52=7$
- At $(0,2)$: $Z=10$
- At $(3,0)$: $Z=9$

The minimum value is $7$ at $(\frac32,\frac12)$.

Constraints:

- $x+2y\le 10$
- $3x+y\le 15$
- $x\ge 0,\ y\ge 0$

Corner points:
- $(0,0)$
- $(5,0)$ from $3x+y=15$ with $y=0$
- $(0,5)$ from $x+2y=10$ with $x=0$
- Intersection of $x+2y=10$ and $3x+y=15$:

From $3x+y=15$, $y=15-3x$.
Substitute in $x+2y=10$:
$x+2(15-3x)=10$
$x+30-6x=10$
$-5x=-20\Rightarrow x=4$.
Then $y=15-12=3$.
So intersection is $(4,3)$.

Now evaluate $Z=3x+2y$:
- $(0,0)$: $0$
- $(5,0)$: $15$
- $(0,5)$: $10$
- $(4,3)$: $12+6=18$

The maximum value is $18$ at $(4,3)$.