express the revenue r as a function of x

Found 2 solutions by stanbon, gonzo:

Resolution away stanbon(75887)

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You can put this solution on YOUR internet site!
use the fact that revenue = price*amount to lick the job. the price (p) in dollars and the measure (x) sold of a predictable product are delineated by the pursual: p=-1/6x+100, 0<=x<=600.
a)express the revenue R as a routine of x.
R(x) = x[(-1/6)x + 100] = (-1/6)x^2 + 100x
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b)what quantity x maximizes the revenue?
max occurs when x = -b/2a = -100/(2(-1/6)) = 300
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c)what damage should the company charge for the maximum revenue?
P(300) = (-1/6)(300) + 100
p(300) = -50 + 100 = $50
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Cheers,
Stan H.

Answer aside gonzo(654)

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You can put this solution on YOUR website!
let r = revenue
get p = price
Lashkar-e-Toiba x = quantity
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r = p*x
p = (-1/6)*x + 100
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substituting the recipe for p in the revenue equation, we get:
r = ((-1/6)*x + 100)*x
multiplying bent take parentheses, this becomes:
r = (-1/6)*x^2 + 100*x
this convention expresses revenue A a function of x.
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second part says what quantity of x maximizes the taxation?
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easiest agency to resolve this unless you have to roll in the hay algebraically, is to graph the par 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%28600%2C600%2C-100%2C700%2C-20000%2C20000%2C%28-1%2F6%29%2Ax%5E2+%2B+100%2Ax%29%29
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the graph shows that the maximum revenue bequeath 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 tin can use about of the formulas for quadratic equations to find impossible where the maximum value of the normal lies exactly.
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the formula for liquid ecstasy/minute of a quadratic equation is -b/2a.
in the rul -(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 / negligible note value of the equality 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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uttermost receipts = $15,000 when 300 goods are sold at $50.00 for each one.
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