SOLUTION: {{{P(x) = -x^3 + (27/2) x^2 - 60x + 100}}}, x ≥ 5 is an approximation of the total profit (in thousands of dollars) from the sale of x hundred thousand tires. Find the number

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: {{{P(x) = -x^3 + (27/2) x^2 - 60x + 100}}}, x ≥ 5 is an approximation of the total profit (in thousands of dollars) from the sale of x hundred thousand tires. Find the number      Log On


   



Question 1041296: P%28x%29+=+-x%5E3+%2B+%2827%2F2%29+x%5E2+-+60x+%2B+100, x ≥ 5 is an approximation of the total profit (in thousands of dollars) from the sale of x hundred thousand tires. Find the number of hundred thousands of tires that must be sold to maximize profit.
I know the answer is 5, but how exactly do I solve this?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
P%28x%29+=+-x%5E3+%2B+%2827%2F2%29+x%5E2+-+60x+%2B+100 ,x ≥ 5
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Is this in a calculus course? If so:
Find the derivative of +P+ with respect to +x+
+P%5B1%5D+=+-3x%5E2+%2B+27x+-+60+
Set the derivative equal to zero
+-3x%5E2+%2B+27x+-+60+=+0+
+-x%5E2+%2B+9x+-+20+=+0+
+%28+-x+%2B+4+%29%2A%28+x+-+5%29+=+0+ ( by looking at it )
+x+=+4+
+x+=+5+
Which one is the maximum?
I know they say +x+%3E=+5+, but I still want
to know if +x+=+5+ is at a maximum
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For that, take the 2nd derivative of +P+
It will be negative for +x%5Bmax%5D+ and positive for +x%5Bmin%5D+
+P%5B2%5D+=+-6x+%2B+27+
+P%284%29+=+-6%2A4+%2B+27+
+P%284%29+=+-24+%2B+27+
+P%284%29+=+3+
and
+P%285%29+=+-6%2A5+%2B+25+
+P%285%29+=+-30+%2B+25+
+P%285%29+=+-5+
Therefore the maximum is at +x+=+5+
500,000 tires must be sold to maximize profit
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