SOLUTION: A company can produce a doll at a cost of $5 per dolls. When the company sells the dolls at a price of $40 they can sell 400 dolls each week. It is estimated that for every $0.50 d

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A company can produce a doll at a cost of $5 per dolls. When the company sells the dolls at a price of $40 they can sell 400 dolls each week. It is estimated that for every $0.50 d      Log On


   



Question 1094144: A company can produce a doll at a cost of $5 per dolls. When the company sells the dolls at a price of $40 they can sell 400 dolls each week. It is estimated that for every $0.50 decrease in the price, they can sell 20 more dolls.
Find the price that will maximize the profit for the company.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +P+ = profit per week
Let +n+ = the number of $.50 decreases in price
[ profit ] = [ income ] - [ cost ]
+P+=+%28+40+-+.5n+%29%2A%28+400+%2B+20n+%29+-+%28+400+%2B+20n+%29%2A5+
+P+=+%28+40+-+5+-+.5n+%29%2A%28+400+%2B+20n+%29+
+P+=+%28+35+-+.5n+%29%2A%28+400+%2B+20n+%29+
+P+=+14000+-+200n+%2B+700n+-+10n%5E2+
+P+=+-10n%5E2+%2B+500n+%2B+14000+
+P+=+-n%5E2+%2B+50n+%2B+1400+
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The n-value of the maximum is at:
+n%5Bmax%5D+=+-b%2F%282a%29+
+a+=+-1+
+b+=+50+
+n%5Bmax%5D+=+-50%2F%282%2A%28-1%29%29+
+n%5Bmax%5D+=+25+
and
+P%5Bmax%5D+=+-10%2A25%5E2+%2B+500%2A25+%2B+14000+
+P%5Bmax%5D+=+-6250+%2B+12500+%2B+14000+
+P%5Bmax%5D+=+20250+
The maximum weekly profit is $20,250
The price that will maximize the profit is:
+40+-+.5n+
+40+-+.5%2A25+
+40+-+12.5+=+27.5+
$27.50 per doll
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check:
Here's the plot:
+graph%28+400%2C+400%2C+-10%2C+60%2C+-2000%2C+25000%2C+-10x%5E2+%2B+500x+%2B+14000+%29+