SOLUTION: Caresia is an artist who works at a her shop drawing “watercolor portraits.” She charges $20 per portrait and she has been averaging 30 portraits per week. She decides to increase

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Question 997186: Caresia is an artist who works at a her shop drawing “watercolor portraits.” She charges $20 per portrait and she has been averaging 30 portraits per week. She decides to increase the price, but realizes that for every one dollar increase she will lose one sale per week. If materials cost her $10 per portrait, what should she set the price at in order to maximize her profit?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +P+ = her profit per week
Let +n+ = number of $1 increases in her price
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+P+=+%28+20+%2B+n+%29%2A%28+30+-+n+%29+-+10%2A%28+30+-+n+%29+
+P+=+%28+20+-+10+%2B+n+%29%2A%28+30+-+n+%29+
+P+=+%28+10+%2B+n+%29%2A%28+30+-+n+%29+
+P+=+300+%2B+30n+-+10n+-+n%5E2+
+P+=+-n%5E2+%2B+20n+%2B+300+
The maximum +n+ is at +-b%2F%282a%29+ where
+a+=+-1+
+b+=+20+
+-b%2F%282a%29+=+-20%2F%282%2A%28-1%29%29+
+-b%2F%282a%29+=+10+
+n+=+10+
Plug this value of n back into the equation
to find the maximum profit
+P+=+-n%5E2+%2B+20n+%2B+300+
+P%5Bmax%5D+=+-10%5E2+%2B+20%2A10+%2B+300+
+P%5Bmax%5D+=+-100+%2B+200+%2B+300+
+P%5Bmax%5D+=+400+
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Her price should be
+20+%2B+n+=++20+%2B+10+
+20+%2B+n+=+30+
$30 per portrait