If cost of producing nn items by a production entity is 950+63n950+63n and price of one item is 280-5n2805n, find the range of nn for which the entity is profitable?

1 Answer
Oct 6, 2017

5<= n <= 2125n212

Explanation:

As cost of producing nn items is 950+63n950+63n

As price of one item (if nn are sold) is 280-5n2805n and

selling price of nn items is 280n-n^2280nn2

Hence for profit, we should have

280n-n^2 >= 950+63n280nn2950+63n

or n^2+63n-280n+950 <= 0n2+63n280n+9500

or n^2-217n+950 <= 0n2217n+9500

i.e. if alphaα and betaβ are roots of n^2-217n+950=0n2217n+950=0 where alpha < betaα<β.

and then (n-alpha)(n-beta)<=0(nα)(nβ)0

Roots of n^2-217n+950=0n2217n+950=0 using quadratic formula are

n=(217+-sqrt(217^2-4xx950))/2n=217±21724×9502

or (217+-208.06)/2217±208.062 i.e. 212.53212.53 or 4.474.47

and so beta=212.53β=212.53 or alpha=4.47α=4.47

As (n-alpha)(n-beta)<=0(nα)(nβ)0

we have either n-alpha>=0nα0 and n-beta<=0nβ0 i.e. n>=alphanα and n<=betanβ i.e. alpha <= n <= betaαnβ

or n-alpha<=0nα0 and n-beta>=0nβ0 i.e. n<=alphanα and n>=betanβ, which is not possible as alpha < betaα<β.

Hence answer is 4.47<= n <= 212.534.47n212.53 but as we should have only integers answer is 5 <= n <= 2125n212