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 the 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 2x, so the total storage and setup cost is C = 500y +2x. Furthermore, xy = 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

Respuesta :

Answer:

B) The company should have 8 production runs each year

Explanation:

Given :

Uniform annual demand, = 16000

Total cost, C = 500y + 2x

xy = 16000

[tex] x = \frac{16000}{y} [/tex]

Let's substitute [tex]\frac{16000}{y}[/tex] for x in C.

Therefore, we have :

[tex] C = 500y + 2( \frac{16000}{y} )[/tex]

[tex] C = 500y + \frac{32000}{y}[/tex]

In order to minimize the total storage and setup costs,

Differentiating wrt y:

[tex] C = C_m_i_n, \frac{dc}{dy}=0[/tex]

[tex] C'(y) = 500y + \frac{32000}{y^2} = 0 [/tex]

[tex] y^2 = \frac{320}{5} = 64 [/tex]

[tex] y = \sqrt{64} = 8 [/tex]

In order to minimize the total storage and setup costs, the company should have 8 production runs each year