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KEAM 2020 Math Paper
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© examsnet.com
Question : 54
Total: 120
The maximum value of
z
=
7
x
+
5
y
subject to
2
x
+
y
≤
100
,
4
x
+
3
y
≤
240
,
x
≥
0
,
y
≥
0
is
350
380
400
410
420
Validate
Solution:
z
=
7
x
+
5.4
(
50
,
0
)
7
×
50
+
0
350
(
0
,
80
)
0+400
400
(
30
,
40
)
210+200
410
© examsnet.com
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