Question

The cost of manufacturing of certain items consists of INR \[1600\] as overheads, INR \[30\] per item as cost of material and the labour costs INR \[\dfrac{{{x^2}}}{{100}}\]​ for \[x\] items produced. How many items must be produced to have a minimum average cost?

Answer 1

Hint: We are given that items are produced with INR \[1600\] as overheads, INR \[30\] as cost of material per item and INR \[\dfrac{{{x^2}}}{{100}}\] as the labour costs. Now we need to find how many items are needed to minimise the average cost. Let know the average cost, primarily find the total cost by adding all the expenses together and later divide the sum by the total product. And finally differentiate the result we have.

Complete step by step answer:
Let us consider the items produced \[ = x\]
Overhead cost \[ = \] Rs \[1600\]
Cost of material per item \[ = \] Rs \[30\]
Therefore, Cost of material \[ = 30x\]
Labour costs \[ = \] Rs \[\dfrac{{{x^2}}}{{100}}\]
Now, Total cost \[ = \] overhead \[ + \] cost of material \[ + \] labour cost
\[\text{Total cost}= 1600 + 30x + \dfrac{{{x^2}}}{{100}}\]
Now divide the whole term by \[x\] to get the average total cost
Average of total cost \[ = \dfrac{{{\text{overhead }} + {\text{cost of material }} + {\text{labour cost }}}}{{{\text{total product }}}}\]
\[\text{Average of total cost} = \dfrac{{1600 + 30x + \dfrac{{{x^2}}}{{100}}}}{x}\]
\[x\] is common for all the terms so take it separately.
\[\text{Average of total cost} = \dfrac{{1600}}{x} + \dfrac{{30x}}{x} + \dfrac{{\dfrac{{{x^2}}}{{100}}}}{x}\]
\[x\] in both the numerator and the denominator of \[\dfrac{{30x}}{x}\] will get cancel and taking the reciprocal of \[x\] in the denominator of \[\dfrac{{{x^2}}}{{100}}\],
\[\text{Average of total cost} = \dfrac{{1600}}{x} + 30 + \dfrac{{{x^2}}}{{100}} \times \dfrac{1}{x}\]
Now a \[x\] in the \[\dfrac{{{x^2}}}{{100}}\] will get cancel by \[\dfrac{1}{x}\]
\[\text{Average of total cost} = \dfrac{{1600}}{x} + 30 + \dfrac{x}{{100}}\]

To know the items produced by the minimum average cost, take differentiation of the average cost and equate it with zero.For minimum of a function, \[\dfrac{{df}}{{dx}} = 0\].
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{1600}}{x} + 30 + \dfrac{x}{{100}}} \right] = 0\]
We have the formulas: \[\dfrac{d}{{dx}}\left( {\dfrac{1}{{{x^n}}}} \right) = - \dfrac{1}{{{x^{n + 1}}}}\]
\[\dfrac{d}{{dx}}({\text{constant}}) = 0\]
\[ \Rightarrow\dfrac{d}{{dx}}\left( {\dfrac{x}{n}} \right) = \dfrac{1}{n}\]
From the formulas,
\[\dfrac{d}{{dx}}\left( {\dfrac{1}{{{x^n}}}} \right) = - \dfrac{1}{{{x^{n + 1}}}} \\
\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{1600}}{x}} \right) = \dfrac{{ - 1600}}{{{x^{1 + 1}}}} = \dfrac{{ - 1600}}{{{x^2}}}\],
\[\dfrac{d}{{dx}}({\text{constant}}) = 0 \\
\Rightarrow \dfrac{d}{{dx}}(30) = 0\] and
\[\dfrac{d}{{dx}}\left( {\dfrac{x}{n}} \right) = \dfrac{1}{n} \\
\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{x}{{100}}} \right) = \dfrac{1}{{100}}\] then,
\[ - \dfrac{{1600}}{{{x^2}}} + 0 + \dfrac{1}{{100}} = 0\]

This can be rearrange and write as,
\[\dfrac{1}{{100}} - \dfrac{{1600}}{{{x^2}}} = 0\]
Take L.C.M. by cross multiplication,
\[\dfrac{{1({x^2}) - 1600(100)}}{{100{x^2}}} = 0\]
\[\dfrac{{{x^2} - 160000}}{{100{x^2}}} = 0\]
Move the denominator to R.H.S.,
\[{x^2} - 160000 = 0 \times 100{x^2}\]
Any number multiplied by \[0 = 0\],
\[{x^2} - 160000 = 0\]
Now, \[{x^2} = 160000\]
\[x = \sqrt {160000} \]
Taking the square root of \[160000\] we will get,
\[\therefore x = 400\]

Therefore, \[400\] items should be produced to minimise the average cost.

Note: We know that differentiation is also used to find the maximum and minimum values of a function. And here the given question is also about the number of products to minimise the average cost. So it becomes necessary to read the question until identifying the hidden details. Then only we can come to a conclusion. In the given question the word minimum drives the whole problem into the path of differentiation.