Question

Minimize $Z = 6x + 3y$ under the constraints: \[ 4x + y \geq 80, x + 5y \geq 115, 3x + 2y \leq 150, x \geq 0, \; y \geq 0. \]

Hint:

In LPP, always evaluate $Z$ at all corner points of the feasible region to find the optimal solution.

Answer 1

This is a linear programming problem (LPP). The feasible region is bounded by the lines: 1. $4x + y = 80 \;\; $\Rightarrow$ \;\; y = 80 - 4x$ 2. $x + 5y = 115 \;\; $\Rightarrow$ \;\; y = \tfrac{115 - x}{5}$ 3. $3x + 2y = 150 \;\; $\Rightarrow$ \;\; y = \tfrac{150 - 3x}{2}$

Step 1: Find corner points of the feasible region. - Intersection of $4x+y=80$ and $x+5y=115$: \[ 4x+y=80 $\Rightarrow$ y=80-4x \] Substitute in $x+5y=115$: \[ x+5(80-4x)=115 \;\; $\Rightarrow$ \;\; x+400-20x=115 \;\; $\Rightarrow$ \;\; -19x=-285 \;\; $\Rightarrow$ \;\; x=15, \; y=20. \] - Intersection of $4x+y=80$ and $3x+2y=150$: \[ y=80-4x, 3x+2(80-4x)=150 \;\; $\Rightarrow$ \;\; 3x+160-8x=150 \] \[ -5x=-10 \;\; $\Rightarrow$ \;\; x=2, \; y=72. \] - Intersection of $x+5y=115$ and $3x+2y=150$: \[ x=115-5y, 3(115-5y)+2y=150 \] \[ 345-15y+2y=150 \;\; $\Rightarrow$ \;\; -13y=-195 \;\; $\Rightarrow$ \;\; y=15, \; x=40. \]

Step 2: Evaluate objective function $Z=6x+3y$ at corner points. - At $(15,20)$: $Z=6(15)+3(20)=90+60=150$ - At $(2,72)$: $Z=6(2)+3(72)=12+216=228$ - At $(40,15)$: $Z=6(40)+3(15)=240+45=285$

Step 3: Find minimum. The minimum value is \[ Z_{\min} = 150 \text{at} \; (15,20). \]

Final Answer: \[ \boxed{Z_{\min}=150 \; \text{at} \; (15,20)} \]