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Standard VI
Mathematics
Formation of Algebraic Equations
A decorative ...
Question
A decorative item dealer deals in two items A and B. he had Rs 15,000 to invest and a space to store at the most 80 pieces. Item A cost him Rs 300 and item B costs him Rs 150 . he can sell item A and B at respective profits of Rs 50 and Rs 28. formulate Lpp in order to maximum.
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Solution
Dear Student,
Let
x
pieces
of
item
A
and
y
pieces
of
item
B
is
sold
.
The
respective
profits
on
item
A
and
item
B
are
Rs
50
and
Rs
28
.
So
,
the
objective
function
is
,
Z
=
50
x
+
28
y
The
total
number
of
pieces
he
can
store
is
80
.
So
,
we
get
x
+
y
≤
80
Also
,
the
total
amount
of
money
he
had
is
Rs
15
,
000
and
per
unit
price
of
item
A
and
item
B
are
Rs
300
and
Rs
150
resepctively
.
This
further
gives
,
300
x
+
150
y
≤
15
,
000
or
2
x
+
y
≤
100
Combining
the
objective
function
and
the
above
constraints
,
we
get
the
required
LPP
as
follows
:
Maximize
Z
=
50
x
+
28
y
Subject
to
,
x
+
y
≤
80
2
x
+
y
≤
100
x
≥
0
,
y
≥
0
Regards
Suggest Corrections
0
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