Maximum profit

Question

Phillip, the proprietor of a vineyard, estimates that the first 10000 bottles of wine produced this season will fetch a profit of $3 per bottle. However, the profit from each bottle beyond 10000 drops by $0.0002 for each additional bottle sold. Assuming at least 10000 bottles of wine are produced and sold, what is the maximum profit? (Round your answer correct to the nearest cent.)
$


What would be the profit/bottle in this case? (Round the number of bottles down to the nearest whole bottle. Round your answer correct to the nearest cent.)
$

Answer 1

P(x) = number of bottles sold times profit per bottle. 

Let x be number of bottles sold greater than 10,000, so total number bottles sold is 10,000+x.  Profit for each bottle will be 3-.0002x.Thus,

P(x)=(10000+x)(3-.0002x). 

Take derivative and set to 0 to get x = 2,500 bottles. Calculate P(2500) to get max profit of $31,250. 

Note, P”(x) is negative so P(x) is concave down. Hence, 2500 is a max. 

Profit per bottle = 31250/12500 = $2.50. Good luck!

Answer 2

x=# of bottles beyond 10,000

(10,000+x)(3-0.0002x)=0

multiply

30,000+3x-2x-0.0002x^2=0

-0.0002x^2+x+30,000=0

this is a parabola that opens downward so it has a maximum point, (h,k) where h=-b/2a and h=x

-1/(-0.0004)=10,000/4=2500

x=2500 more bottles to be sold

maximum profit: (10,000+2500)(3-0.0002*2500)

(12,500)(3-0.50)

(12,500)(2.50)=$31,250 maximum profit

note: 2501, 2502, 2499, 2498 and a few other numbers will give the same maximum profit because of the very small drop in price of $0.0002 per bottle sold

profit/bottle=31,250/12,500=$2.50 as shown above