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.
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