A paint manufacturer has a uniform annual demand for 16,000 cans of automobile primer. It costs $4 to store one can of paint for one year and $500 to set up the plan for production of the primer. Let x be the number of cans of paint produced during each production run, and let y be the number of production runs. Then the setup cost is 500y and the storage cost is 2.c, so the total storage and setup cost is C = 500y +2.c. Furthermore, .cy = 16,000 to account for the annual demand. How many times a year should the company produce this primer in order to minimize the total storage and setup costs? A. The company should have 6 production runs each year.
B. The company should have 8 production runs each year.
C. The company should have 10 production runs each year.
D. The company should have 11 production runs each year.

B. The company should have 8 production runs each year.

Step-by-step explanation:

From the given information:

A paint manufacturer has a uniform annual demand for 16,000 cans of automobile primer

It costs $4 to store one can of paint for one year

$500 to set up the plan for production of the primer

Let x be the number of cans of paint produced during each production run

Let y be the number of production runs.

If the total storage and setup cost is C = 500y + 2c

and cy = 16000

Then c = 16000/y

From;

C = 500y + 2c

Replacing c with 16000/y,  we have;

C = 500y + 2(16000/y)

C = 500y + 32000/y

in order to minimize the total storage and setup costs  

Therefore

⇒ 500 - 32000/y² =0

y² = 32000/500

y² = 320/5

y² = 64

y = (√64)

y = 8

Therefore; The company should have 8 production runs each year in order to minimize the total storage and setup costs

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%