Question 1204231
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Based on this previous problem here
<a href = "https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1177138.html">https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1177138.html</a>
It appears there should be the instruction that "<font color=red>x hundred bicycles are built</font>"


x = number of bikes in hundreds
eg: x = 3 means 300 bikes are made
C(x) = 0.1x^2-0.7x+2.425 is the cost in hundreds of dollars.


{{{(C(x))/x = (0.1x^2-0.7x+2.425)/x}}} represents the average cost, in hundreds of dollars, of making x hundred bikes.


Use calculus or a graphing calculator to determine the local min of {{{(C(x))/x}}}
GeoGebra and Desmos are graphing tools I recommend.
If you have a TI83 or TI84, then that works as well.


When x > 0, the lowest point happens at roughly (4.92443, 0.28489)


x = 4.92443 leads to 100*4.92443 = 492.443


Of course it's not possible to build 0.443 of a bike. 
We look at integer values to the left and right of 492.443
We'll look at integer values 492 and 493 which correspond to x = 4.92 and x = 4.93 respectively.


If x = 4.92, then the average cost is:
{{{(C(x))/x = (0.1x^2-0.7x+2.425)/x}}}


{{{(C(4.92))/(4.92) = (0.1(4.92)^2-0.7(4.92)+2.425)/(4.92)}}}


{{{(C(4.92))/(4.92) = 0.28488617886179}}} approximately


{{{(C(4.92))/(4.92) = 0.2849}}} approximately


Since {{{(C(x))/x}}} is the average cost in hundreds of dollars, it means the result 0.2849 represents 0.2849*100 = 28.49 dollars.
It costs around 28.49 dollars per bike, on average, if 492 bikes are made.


Now try x = 4.93
{{{(C(x))/x = (0.1x^2-0.7x+2.425)/x}}}


{{{(C(4.93))/(4.93) = (0.1(4.93)^2-0.7(4.93)+2.425)/(4.93)}}}


{{{(C(4.93))/(4.93) = 0.2848864097363}}} approximately


{{{(C(4.93))/(4.93) = 0.2849}}} approximately
We get the same average cost as before. 


It is up to the company to decide whether to make 492 bikes or 493 bikes. 
Either option produces the lowest average cost of $28.49 dollars per bike.
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