Use duality to solve the problem.
-Minimize $w=y_{1}+4 y_{2}+3 y_{3}$
Subject to: $3 y_{1}+2 y_{2}+y_{3} \geq 70$
$2 \mathrm{y}_{1}+2 \mathrm{y}_{3} \geq 50$
$\mathrm{y}_{1} \geq 0, \mathrm{y}_{2} \geq 0, \mathrm{y}_{3} \geq 0$
A) 25
B) 20
C) 90
D) 165

Question

Quizplus · Accepted Answer

The correct answer is A because the dual problem is to maximize $z=70 u_{1}+50 u_{2}$ subject to $3 u_{1}+2 u_{2} \leq 1$, $2 u_{1} \leq 4$, and $u_{1}, u_{2} \geq 0$. The optimal solution to the dual problem is $u_{1}=25$ and $u_{2}=0$, which gives $w=70(25)+50(0)=1750$.

Use Duality to Solve the Problem w=y1+4y2+3y3w=y_{1}+4 y_{2}+3 y_{3}w=y1​+4y2​+3y3​ Subject To 3y1+2y2+y3≥703 y_{1}+2 y_{2}+y_{3} \geq 703y1​+2y2​+y3​≥70

Use Duality to Solve the Problem $w=y_{1}+4 y_{2}+3 y_{3}$ Subject To $3 y_{1}+2 y_{2}+y_{3} \geq 70$