He also supposes that the total cost of keeping n Tvs in inventory is by c(n) = o,3n^2 + 5000.

how many TVs must retailer keep in inventory and sell to maximize his profit?

600

500

750

450

profit = (600-0.3n) *n - (0.3n^2+5000)

= 600n-0.3n^2-0.3n^2-5000

=-0.6n^2+600n-5000

d(profit)/dn = -1.2n + 600 = 0

n = 500

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