Answer:
a) Cost
![C(q) = 50+16q\\\\C(500)=50+16(500)=50+8,000=8,050](https://img.qammunity.org/2021/formulas/mathematics/high-school/llgzwrrex781oeh2kstlp0i15u3kiwm3g0.png)
b) Sales income
![S(q)=20q\\\\S(500)=20\cdot 500 = 10,000](https://img.qammunity.org/2021/formulas/mathematics/high-school/5nlpe9arvfc9sj8ol4i1ymzc2gwufh1wxu.png)
c) Table of values
![\left[\begin{array}{ccc}q&C(q)&S(q)\\0&50&0\\250&4,050&5,000\\500&8,050&10,000\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/high-school/8xu7ew0bk8dqzaqovk02y2jnzcxtr0g7sn.png)
d) Attached
e) Breakeven point = 12.5 sheets
f) Profit at 550 sheets = $1,950
Explanation:
a) We have a fixed cost for the image, at $50.
We also have a variable cost of $16 a sheet.
The purchased quantity is 500 sheets.
Then, the cost function is:
![C(q) = 50+16q\\\\C(500)=50+16(500)=50+8,000=8,050](https://img.qammunity.org/2021/formulas/mathematics/high-school/llgzwrrex781oeh2kstlp0i15u3kiwm3g0.png)
b) The price for each sheet is $20, so the income from sales are:
![S(q)=20q\\\\S(500)=20\cdot 500 = 10,000](https://img.qammunity.org/2021/formulas/mathematics/high-school/5nlpe9arvfc9sj8ol4i1ymzc2gwufh1wxu.png)
c) Table of values
![\left[\begin{array}{ccc}q&C(q)&S(q)\\0&50&0\\250&4,050&5,000\\500&8,050&10,000\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/high-school/8xu7ew0bk8dqzaqovk02y2jnzcxtr0g7sn.png)
d) Attached
e) The minimum number of sheets the group must sell so they don't lose any money is the breakeven point (BEP) and can be calculated making the income sales equal to the cost:
![S(q)=C(q)\\\\20q=50+16q\\\\(20-16)q=50\\\\4q=50\\\\q=50/4=12.5](https://img.qammunity.org/2021/formulas/mathematics/high-school/8wtt8k7i50fyj78zsd6vbcvq7dssbcdp6n.png)
f) This profit can be calculated as the difference between the sales income and the cost:
![P(500)=S(500)-C(500)\\\\P(500)=20\cdot 500-(50+16\cdot 500)=10,000-8,050=1,950](https://img.qammunity.org/2021/formulas/mathematics/high-school/oqvzk8e2iwt9cbqvipdvpeed25nanqzkc3.png)