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

1 Answer
Oct 6, 2017

#5<= n <= 212#

Explanation:

As cost of producing #n# items is #950+63n#

As price of one item (if #n# are sold) is #280-5n# and

selling price of #n# items is #280n-n^2#

Hence for profit, we should have

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

or #n^2+63n-280n+950 <= 0#

or #n^2-217n+950 <= 0#

i.e. if #alpha# and #beta# are roots of #n^2-217n+950=0# where #alpha < beta#.

and then #(n-alpha)(n-beta)<=0#

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

#n=(217+-sqrt(217^2-4xx950))/2#

or #(217+-208.06)/2# i.e. #212.53# or #4.47#

and so #beta=212.53# or #alpha=4.47#

As #(n-alpha)(n-beta)<=0#

we have either #n-alpha>=0# and #n-beta<=0# i.e. #n>=alpha# and #n<=beta# i.e. #alpha <= n <= beta#

or #n-alpha<=0# and #n-beta>=0# i.e. #n<=alpha# and #n>=beta#, which is not possible as #alpha < beta#.

Hence answer is #4.47<= n <= 212.53# but as we should have only integers answer is #5 <= n <= 212#