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Mashcka [7]
3 years ago
11

A start up publishing company estimates that the fixed costs of its first major project will be $190,000, the variable cost will

be $18, and the selling price per book will be $34. A.) how many books must be sold for this project to break even? B.) suppose the publishers wish to take a total of $40,000 in salary for this project. How many books must be sold to break even, and what is the break-even point, in dollars?
Mathematics
1 answer:
Tanya [424]3 years ago
5 0

Solution:

we are given that

Fixed cost=$190,000.

Variable cost = $18.

selling price per book =$34.

(A)

For break even , let the number of books required to sell be x

Then we can write

190000+18x=34x\\\\\text{Solve for x}\\\\190000=34x-18x\\\\16x=190000\\\\x=11875

Hence the number of books required to sell is 11875 for break even.

(B)

For break even , let the number of books required to sell be x

Then we can write

190000+40000+18x=34x\\\\\text{Solve for x}\\\\230000=34x-18x\\\\16x=230000\\\\x=14375

Hence the number of books required to sell is 14375 for break even.

Break even points in dollar is 14375x34=$488750.

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Answer:

\large\boxed{1.\ a_{44}=481,\ \sum\limits_{n=1}^\infty(11n-3),\ \text{the sum not exist}}

\large\boxed{3.\ \sum\limits_{n=1}^{47}(9n+16)=10,904}

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1.\\8+19+30+41+...\\\\19-8=11\\30-19=11\\41-30=11\\\\\text{It's an arithmetic series.}\ a_1=8,\ d=11.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=8+(n-1)(11)\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\a_n=8+11n-11\\\boxed{a_n=11n-3}\\\\\text{Calculate the 44th term. Put n = 44 to the formula:}\\\\a_{44}=(11)(44)-3=484-3=481

\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty(11n-3)\\\\\text{The sum not exist, because}\ d>1,\ \text{therefore}\ 11n-3\rightarrow\infty.

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3.\\25+34+43+52+...+436\\\\34-25=9\\43-34=9\\52-43=9\\\\\text{It's an arithmetic series.}\ a_1=25,\ d=9.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=25+(n-1)(9)\\a_n=25+9n-9\\a_n=9n+16\\\\\text{Calculate which the term of arithmetic series is the number 439.}\\\text{Put}\ a_n=439:\\\\9n+16=439\qquad\text{subtract 16 from both sides}\\9n=423\qquad\text{divide both sides by 9}\\n=47

\text{Therefore we have the series in summation notation:}\\\\\sum\limits_{n=1}^{47}(9n+16)\\\\\text{For calculation of the sum we use the formula of a sum of terms}\\\text{of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=47,\ a_1=25,\ a_{47}=439:\\\\S_{47}=\dfrac{25+439}{2}\cdot47=\dfrac{464}{2}\cdot47=232\cdot47=10904\\\\\sum\limits_{n=1}^{47}(9n+16)=10,904

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4.\\12+24+48+...\\\\24:12=2\\48:24=2\\\\\text{It's\ a\ geometric series}\ a_1=12,\ r=2.\\\\\text{The formula for the n-th term of a geometic sequence:}\\\\a_n=a_1r^{n-1}\\\\\text{Substitute:}\\\\a_n=(12)(2^{n-1})\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\a_n=(12)\left(\dfrac{2^n}{2^1}\right)\\\\a_n=(6)(2^n)\\\\\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg)\\\\\text{The sum not exist, because}\ r>1,\ \text{therefore}\ 6(2^n)\to\infty

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