SOLUTION: an art museum charges $10 for an adult admission. The museum estimates they will lose 15 adult visitors per day for each one dollar increase in price. If the museum averages 300

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Question 453695: an art museum charges $10 for an adult admission. The museum estimates they will lose 15 adult visitors per day for each one dollar increase in price. If the museum averages 300 adult visitors a day, which is the most profitable adult admission price?
Found 2 solutions by htmentor, josmiceli:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
an art museum charges $10 for an adult admission. The museum estimates they will lose 15 adult visitors per day for each one dollar increase in price. If the museum averages 300 adult visitors a day, which is the most profitable adult admission price?
Let x = the admission price
Then the number of visitors will be 300 - 15(x-10) [300 for x=10, 285 for x=11, etc.]
The total revenue will be:
R = x(300 - 15(x-10))
R = 300x - 15x^2 + 150x
R = 450x - 15x^2
To maximize the profit, we set dR/dx = 0 and solve for x:
dR/dx = 0 = 450 - 30x
This gives x = 15
So the most profitable price is $15
The graph of the revenue function is below:
graph%28300%2C300%2C-20%2C20%2C-200%2C3600%2C450x-15x%5E2%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +M+ = money taken in
Let +i+ = number of $1 increases
Money in = (number of adult visitors) x (admission per visitor)
+M+=+%28+300+-+15i+%29%2A%28+10+%2B+i+%29+
+M+=+3000+-+150i+%2B+300i+-+15i%5E2+
+M+=+-15i%5E2+%2B+150i+%2B+3000+
+M+=+-i%5E2+%2B+10i+%2B+200+
The maximum money in occurs at +i+=+-b%2F%282a%29+
where the equation has the form +f%28x%29+=+a%2Ax%5E2+%2B+bx+%2B+c+
+a+=+-1+
+b+=+10+
+i%5Bmax%5D+=+-10%2F%282%2A%28-1%29%29+
+i%5Bmax%5D+=+5+
+10+%2B+i+=+10+%2B+5+
+10+%2B+i+=+15+
The most profitable admission price is $15/visitor