Question

Sheet A computer store sells 7,200 boxes of storage drives annually. It costs the store $0.20 to store a box of drives for one year. Each time it reorders drives, the store must pay a $5.00 service charge for processing the order. 7200 boxes of storage drives are sold every year 0.2 is the cost of storing one box 5 is the cost of ordering boxes X is the number of boxes ordered is the number of orders per year How can you express the boxes of storage drives sold every year in terms of x and y? 7200 The average number of boxes stored every year is x/2. What is the cost function in terms of x and y? In terms of x? Cix.y = C(x) = What is the marginal cost function? Cix) What is the minimum cost? How many orders should the store make every year? Use formulas where you can. X Minimum cost Number of orders:

Answer 1

Solution-

We have given that

N = 7200 boxes of storage drive are sold every year.

C_s = $ 0.2 is the cost of storing one box

C_o = $ 5 is the cost of ordering boxes

x = the number of boxes ordered

y = the number of order per year

(a)

The boxes of storage drives sold every year in terms of x and y can be expressed as

7200 = (Number of boxs ordered)×(number of order per year)

7200 =xy

Hence, 7200 = xy

This gives y = 7200/x ....(1)

(b)

If the average number of boxes stored every year is x/2.

Then the cost function in terms of x and y is

C(x, y) =Co×(y) + C_s×(x/2)

C(x, y) =5y + 0.2x/2

C(x, y) = 5y + 0.1x

So,

C(x, y) = 5y + 0.1x

On putting the value of y from equation (1), we get

C(x) = 5(7200/x) = 0.1x

C(x) = 36000/x + 0.1x ....(2)

(c)

Marginal cost function is the derivative of cost function.

So,

Marginal cost C(x)

36000 +0.17 dz

$=- \frac{36000}{x^2}+0.1(1)$

$=- \frac{36000}{x^2}+0.1$

$=0.1- \frac{36000}{x^2}$

Hence, Marginal cost C(x) = 0.1 - 36000/x².

(d)

For minimum cost, derivative of cost function must be 0.So,

$0.1- \frac{36000}{x^2}=0$

$0.1= \frac{36000}{x^2}$

$x^2= \frac{36000}{0.1}$

x² = 360,000

This implies x = 600

Hence, for minimum cost

Number of boxes ordered at a time = x = 600

Minimum cost is (using equation 2)

​​​​​​C(600) =36000/600 + 0.1(600)

C(600) = 60 + 60 = $ 120

Number of orders per year (using equation 1)

y =7200/x = 7200/600 = 12