Answer:
![C(p)=50p+400](https://img.qammunity.org/2021/formulas/mathematics/high-school/cxfe03v9s24lzedjanb6ysdbkzjifv82em.png)
Explanation:
Let p represent number of phones produced in one day.
We have been given that Allison must pay a daily fixed cost to rent the building and equipment, and also pays a cost per phone produced for materials and labor. The daily fixed costs are $400 and and the total cost of producing 3 phones in a day would be $550.
Let x represent cost of each phone.
We can represent this information in an equation by equating total cost of 3 phones with 550 as:
![3x+400=550](https://img.qammunity.org/2021/formulas/mathematics/high-school/oc0n7398p8jktigsu1kft5jjtv4rs9taol.png)
Let us find cost of each phone.
![3x=550-400](https://img.qammunity.org/2021/formulas/mathematics/high-school/xkxog0rmw3cyus1sxhea1bhalwlsq1mj46.png)
![3x=150](https://img.qammunity.org/2021/formulas/mathematics/high-school/ksr12c8up494kwugido7ryv84j62d23m3c.png)
![x=(150)/(3)](https://img.qammunity.org/2021/formulas/mathematics/high-school/puksmblv0748rq48kk14w2p2tnlku2qiav.png)
![x=50](https://img.qammunity.org/2021/formulas/mathematics/middle-school/yglaf3tevccx4ke49yr22xncy14u04yts1.png)
Since cost of each phone is $50, so cost of p phones would be
.
The total cost would be equal to cost of p phones plus fixed cost.
![C(p)=50p+400](https://img.qammunity.org/2021/formulas/mathematics/high-school/cxfe03v9s24lzedjanb6ysdbkzjifv82em.png)
Therefore, our required cost function would be
.