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Mathematics
Problem
A factory’s daily adjustment cost (in rupees) for producing
n
n
n
units beyond a target of
100
100
100
units is given by
C
(
n
)
=
∣
n
−
100
∣
+
0.01
(
n
−
100
)
2
+
2500
,
C(n)=|n-100| + 0.01(n-100)^2 + 2500,
C
(
n
)
=
∣
n
−
100∣
+
0.01
(
n
−
100
)
2
+
2500
,
where
0
≤
n
≤
200
0\le n\le 200
0
≤
n
≤
200
. Determine the value of
n
n
n
that minimises this cost.
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