Acamera manufacturer spends $1500 each day for overhead expenses plus $9 per camera for labor and materials. the cameras sell for $15 each. how many cameras must the company sell in one day to equal its daily cost? if the manufacturer can increase production by 50 cameras per day, what would their daily profit be?

250 cameras / day

The daily profit would be $300

Step-by-step explanation:

Modeling With Equations

Some situations in real life are adequate for being modeled as functions of the variables they depend on. Equations can be of great help for economy calculations since we could determine optimum levels of production, revenue, costs, and other useful information.

The fixed cost our camera manufacturer is $1500 each day and a variable cost of $9 per camera sold. It can be written as

C(x)=1500+9x

Being C(x) the total cost of manufacturing, and x the number of cameras sold each day

If the company sells the cameras for $15 apiece, then the revenue for x cameras will be

R(x)=15x

a)

We want to find out how many cameras must be sold to equal its daily cost, so

1500+9x=15x

6x=1500

x=250 cameras / day

b) Given the manufacturer can increase production by 50 cameras per day (x=300), then the revenue will be

R(300)=15(300)=$4500

And the cost

C(300)=1500+9(300)=4200

The daily profit would be $4500-$4200=$300

i think 5 miles per hour

step-by-step explanation:

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives   -2, 0, 2, 3.

you,

mrb