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Basic algebra: find breakeven point 
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#1
Jan1613, 04:34 PM

P: 108

1. The problem statement, all variables and given/known data
Mike make shirts. He has fixed daily costs of $150. It costs an additional $3 to make each shirt. He would like to make a profit of $750 a day making shirts. If he can make 24 shirts a day, how much must he charge to meet his goal? Find break even point. 2. Relevant equations c(x) = 3x + 150 p = r(x)  c(x) 3. The attempt at a solution 1) p = r(x)  c(x) p = r(x)  c(24) 750 = r(x)  222 750 = 972  222 972 = r(24) r = 40.5 r(x) = 40.5x He must charge $40.50. 2)40.5x = 3x + 150 37.5x = 150 x = 4 Break even point is $4 


#2
Jan1613, 04:47 PM

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It's not clear whether the last part means a breakeven price or a breakeven output. In your equations you've taken it to be breakeven output, so the answer should be a number of shirts, not a number of dollars. And it's exact, not approximate.



#3
Jan1613, 05:34 PM

P: 108

Btw, what if for a break even point, the "price" gives something approximate, i.e. 350.55 shirts for one and 350.71 for the other? Would you just take the break even point to be the first price that gives more revenue than cost? 


#4
Jan1613, 07:23 PM

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Basic algebra: find breakeven point



#5
Jan1713, 08:21 AM

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The "break even point" is NOT a price. It is the number of shirts he must sell in order to just meet his costs.
You assumed that yourself when you wrote "40.5x = 3x + 150". $40.50 is the price he is getting for each shirt. so 40.5x is the gross income if x is the number of shirts. Similarly, $3 is the marginal cost of each shirt so 3x is a cost only if x is the number of shirts. Your answer should be "He must make 4 shirts a day to break even", NOT "$4". 


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