Question 167650
let r = revenue
let p = price
let x = quantity
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r = p*x
p = (-1/6)*x + 100
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substituting the formula for p in the revenue equation, we get:
r = ((-1/6)*x + 100)*x
multiplying out to remove parentheses, this becomes:
r = (-1/6)*x^2 + 100*x
this formula expresses revenue as a function of x.
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second part says what quantity of x maximizes the revenue?
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easiest way to solve this unless you have to do it algebraically, is to graph the equation and see where the maximum lies.
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graph of the equation looks like this:
look below the graph for further comments.
{{{graph(600,600,-100,700,-20000,20000,(-1/6)*x^2 + 100*x))}}}
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the graph shows that the maximum revenue will occur when x is around 300 and that the maximum revenue will be around 15000.
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since the graph is a quadratic, we can use some of the formulas for quadratic equations to find out where the maximum value of the formula lies exactly.
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the formula for max/min of a quadratic equation is -b/2a.
in the formula -(1/6)*x^2 + 100*x,
a = (-1/6), and
b = 100
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-b/2a = -100/(-1/3)
this is the same as:
-100*(-3) which equals 300.
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the maximum / minimum value of the equation occurs at x = 300.
since the a value is negative, the equation opens downward and points upward, so the value of x = 300 is a maximum value.
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when x = 300, the value of r = (-1/6)*(300)^2 + 100*300.
this becomes:
(-1/6)*90000 + 30000
this becomes:
-15000 + 30000
this becomes:
15000
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price = (-1/6)*300 + 100 = $50.00 per goods sold.
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maximum revenue = $15,000 when 300 goods are sold at $50.00 apiece.
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