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?

a)The company has to 250 cameras in one day to equal its daily cost

b)Their daily profit  will be $400

Step-by-step explanation:

Given:

overhead expenses = $1500

labour and materials expenses = $9

cameras sell for $15 each

To Find:

(a)How many cameras must the company sell in one day to equal its daily cost

(b) If the manufacturer can increase production by 50 cameras per day, what would their daily profit be?

Solution:

a) let the number of cameras sold per day be x

Total daily cost = overhead expenses +labour and materials expenses

Total daily cost = 1500 + 9x--------------------------(1)

Total daily revenue = 15x------------------------------(2)

Equating (1) and (2)

1500 + 9x = 15x

1500 = 15x- 9x

1500 = 6x

x=

x= 250

So, the company has to 250 cameras in one day to equal its daily cost.

b) increasing production by 50 cameras per day will give a daily profit of

=>50 X (17-9)

=>50 X 8

=> 400

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )