A soft-drink vendor at a popular beach analyzes his sales records and finds that

label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

P(x) = 0.001x2 + 3x − 1825.

What is his maximum profit per day?
 

How many cans must he sell for maximum profit?
 cans
Mar 10th, 2015

P(x) = −0.001x2 + 3x − 1825.

P(x) = -0.001 (x2-3000x) -1825

P(X) = -0.001 (x-1500)2+ 15002*0.001-1825

P(X) = -0.001 (x-1500)2+ 2250-1825

P(X) = -0.001 (x-1500)2+425

It will have maximum at x=1500 . Maximum value =425

1500 cans 

Max value =425

Mar 10th, 2015

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