Question

$Z=7x+y$, subject to $5x+y\ge 5,x+y\ge 3,x\ge 0,y\ge 0$. The minimum value of Z occurs at

• A. $(3,0)$
• B. $\left(\frac{1}{2},\frac{5}{2}\right)$
• C. $(7,0)$
• D. $(0,5)$

Answer 1

Answer: (D)

We have, maximize $Z=7x+y$,
Subject to :
$5x+y\ge 5,x+y\ge 3,x,y\ge 0$.
Let ${\ell }_1:5x+y=5$
${\ell }_2:x+y=3$
${\ell }_3:x=0$ and ${\ell }_4:y=0$
Shaded portion is the feasible region,
Where $A\left(3,0\right),B\left(\frac{1}{2},\frac{5}{2}\right),C\left(0,5\right)$

For B : Solving ${\ell }_1$ and ${\ell }_2$, we get $B\left(\frac{1}{2},\frac{5}{2}\right)$
Now maximize $Z=7x+y$
Z at $A\left(3,0\right)=7\left(3\right)+0=21$
Z at $B\left(\frac{1}{2},\frac{5}{2}\right)=7\left(\frac{1}{2}\right)+\frac{5}{2}=6$
Z at $C\left(0,5\right)=7\left(0\right)+5=5$
Thus Z, is minimized at $C\left(0,5\right)$ and its minimum value is $5$