Determine the maximum value of $Z=3x+4y$ if the feasible region (shaded) for a $LPP$ is shown in the adjacent figure.
(N/A) The lines $x+2y=76$ and $2x+y=104$ intersect at point $E$. To find the intersection,we solve the system:
$x+2y=76$ $(1)$
$2x+y=104$ $(2)$
From $(1)$,$x=76-2y$. Substituting into $(2)$:
$2(76-2y)+y=104$
$152-4y+y=104$
$-3y=-48 \implies y=16$
Then $x=76-2(16)=76-32=44$.
So,the intersection point is $E(44, 16)$.
From the graph,the corner points of the bounded feasible region are $O(0,0)$,$A(52,0)$,$E(44,16)$,and $D(0,38)$.
We evaluate $Z=3x+4y$ at each corner point:
Comparing these values,the maximum value of $Z$ is $196$ at the point $(44,16)$.
$x+2y=76$ $(1)$
$2x+y=104$ $(2)$
From $(1)$,$x=76-2y$. Substituting into $(2)$:
$2(76-2y)+y=104$
$152-4y+y=104$
$-3y=-48 \implies y=16$
Then $x=76-2(16)=76-32=44$.
So,the intersection point is $E(44, 16)$.
From the graph,the corner points of the bounded feasible region are $O(0,0)$,$A(52,0)$,$E(44,16)$,and $D(0,38)$.
We evaluate $Z=3x+4y$ at each corner point:
| Corner Point | Value of $Z=3x+4y$ |
| $(0,0)$ | $3(0)+4(0)=0$ |
| $(52,0)$ | $3(52)+4(0)=156$ |
| $(44,16)$ | $3(44)+4(16)=132+64=196$ |
| $(0,38)$ | $3(0)+4(38)=152$ |
Comparing these values,the maximum value of $Z$ is $196$ at the point $(44,16)$.
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