The cost to manufacture a product is proportional to the quantity produced, with a cost of $35000 dollars per day when 700 items are produced. The cost to store n items is proportional to the square of n, with a daily storage cost of $3430 dollars when 700 items are stored. In addition, there is a fixed daily cost of $7000. The total daily cost, T when producing and storing n items is the sum of production, storage and fixed costs.

Write a formula for T as a function of n.

Respuesta :

Solution :

[tex]$T_1$[/tex] = k.n (cost proportional to quantity)

35000 = k . 700

k = [tex]$\frac{35000}{700}$[/tex]

  = 50

So [tex]$T_1$[/tex] = 50 n

Now [tex]$T_2=Pn^2$[/tex]     (cost proportional to square of n)

[tex]$3430= P(700)^2$[/tex]

3430 = 490000 P

[tex]$P = \frac{3430}{490000}$[/tex]

[tex]$P = \frac{7}{1000}$[/tex]

[tex]$T_2=\frac{7}{1000}n^2$[/tex]

Now, [tex]$T_3= 7000$[/tex]   (fixed)

Therefore,

Total, [tex]$T=T_1+T_2+T_3$[/tex]

[tex]$T=50n+\frac{7}{1000}n^2+7000$[/tex]

[tex]$T=0.007n^2+50n+7000$[/tex]

The required formula in terms of T and n is T = 0.000[tex]n^{2}[/tex] + 50 n +7000.

Given that ,

The total daily cost = T

When producing and storing n items is the sum of production, storage and fixed costs.

The manufacturing cost of a product when 700 items produced = $35000 dollars per day.

Daily storage cost of when 700 items are stored = $3430 dollars

We have write the formula in terms of total daily cost T and n stored items.

According to the question,

The cost to manufacture a product is proportional to the quantity produced, with a cost of $35000 dollars per day when 700 items are produced.

Where ,total daily cost T and n stored items.

[tex]T_1 = k . n[/tex]

$35000 =  k.700

[tex]k = \frac{35000}{700}[/tex]

k = $50

The cost to store n items is proportional to the square of n, with a daily storage cost of $3430 dollars when 700 items are stored.

[tex]T_2 = pn^{2}[/tex]

$3430 = P [tex](700)^{2}[/tex]

$3430 = P 490000

[tex]P = \frac{3430}{490000} \\\\P = \frac{7}{1000}[/tex]

There is a fixed daily cost of $7000.

[tex]T_3 = 7000[/tex]

Combine all the equation ,

Where T = [tex]T_1 = T_2 = T_3[/tex] and p = k =n

T = 0.000[tex]n^{2}[/tex] + 50 n +7000

Hence,  The required formula in terms of T and n is T = 0.000[tex]n^{2}[/tex] + 50 n +7000.

For the more information about  System of equations click the link given below.https://brainly.com/question/14418346

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