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unit cost for producing toasters #201

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I think part "A" is pretty straightforward. It is the area under the C'(x) curve constrained by the x-values
of 0 and 300 plus the daily $800 fixed cost:


 

= $800 + [0.0003x
3
/3 - 0.18x
2
/2 + 22x]
NOTE: the integral=0 when x=0 so we don't need to include the x=0 case for part "A".
= $800+(0.0003)(300)(300)(300)/3 - 0.18(300)(300)/2 + 22(300) = 800+2700-8100+6600=$2000
This means the unit cost for producing 300 toasters in a single day (including the $800 fixed daily cost) is
$2000/300=$6.67.
For part "B" the integral is re-evaluated over the range of 201 to 300--but what to do about the $800
fixed daily cost? I don't think theproblem statement makes this clear. SInce the fixed cost (overhead) is
$800 whether 0, 1, 200 or 300 toasters are made, I choose to not include that cost in the manufacture of
toasters 201-300. Re-evaluaing the integral over the range 201 to 300:
= [(0.0003)(300)(300)(300)/3-0.18(300)(300)/2+22(300)]-[(0.0003)(201)(201)(201)/3-0.18(201)(201)/2 +
22(201)=[2700-8100+6600]-[812.06-3636.09+4422]=$1200-2408=$1692
This means the unit cost for producing toasters #201 through 300 (not including the $800 fixed daily
cost) is $1692/100=$16.92.

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I think part "A" is pretty straightforward. It is the area under the C'(x) curve constrained by the x-values of 0 and 300 plus the daily $800 fixed cost: = $800 + [0.0003x3/3 - 0.18x2/2 + 22x] NOTE: the integral=0 when x=0 so we don't need to include the x=0 case for part "A". = $800+(0.0003) ...
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