Question

Maximize $Z = 8000x + 12000y$ under the constraints: \[ 9x + 12y \leq 180, 3x + 4y \leq 60, x + 3y \leq 30, x \geq 0, \; y \geq 0 \]

Hint:

In LPP, the maximum/minimum value always occurs at a corner point of the feasible region.

Answer 1

Step 1: Simplify constraints.
\[ 9x + 12y \leq 180 \;\;\Rightarrow\;\; 3x + 4y \leq 60 (\text{same as 2nd constraint}) \] So effective constraints are: \[ 3x + 4y \leq 60, x + 3y \leq 30, x \geq 0, \; y \geq 0 \]

Step 2: Find corner points.
- When $x=0$: $4y \leq 60 $\Rightarrow$ y \leq 15$, and $3y \leq 30 $\Rightarrow$ y \leq 10$. So feasible point $(0,10)$. - When $y=0$: $3x \leq 60 $\Rightarrow$ x \leq 20$, and $x \leq 30$. So feasible point $(20,0)$. - Intersection of $3x+4y=60$ and $x+3y=30$: \[ 3x+4y=60 (1), x+3y=30 (2) \] From (2), $x=30-3y$. Substitute into (1): \[ 3(30-3y) + 4y = 60 \;\;\Rightarrow\;\; 90-9y+4y=60 \;\;\Rightarrow\;\; -5y=-30 \;\;\Rightarrow\;\; y=6, \; x=12 \] So intersection point $(12,6)$.

Step 3: Evaluate objective function at feasible points.
\[ Z = 8000x+12000y \] - At $(0,0)$: $Z=0$ - At $(20,0)$: $Z=160000$ - At $(0,10)$: $Z=120000$ - At $(12,6)$: $Z=8000(12)+12000(6)=96000+72000=168000$

Step 4: Conclusion.
Maximum value of $Z$ is \[ \boxed{168000 \;\;\text{at}\;\; (x,y)=(12,6)} \]