Maximum profit, given revenue and cost equations. A small company produces and sells x products per week. They find that their cost in dollars is C(x) = 50 + 3x and their revenue is R(x) = 6x - 1/100 x^2 . How many products per week should be produced for maximum profit? How would I solve this algebraically (not the calculus way)?
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Expert Answers
calendarEducator since 2011
write5,348 answers
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You should use the equation that relates profit, cost and revenue such that:
`P(x) = R(x) - C(x)`
`P(x)` represents the profit function
`R(x)` represents the revenue function
`C(x)` represents the cost function
Substituting `6x - 1/100 x^2` for `R(x)` and `50 + 3x` for `C(x)` yields:
`P(x) = 6x - 1/100 x^2 - 50 - 3x`
You need to combine the terms that contain the same powers of x such that:
`P(x) = - 1/100 x^2 + 3x - 50`
You need to notice that the equation of provit function is a quadratic equation and since the leading coefficient is negative, the maximum value of the profit function `y = (4ac - b^2)/(4a)` is reached at `x = -b/(2a).`
Identifying the coefficients a,b,c yields:
`a = -1/100, b = 3, c = -50`
`x = -3/(-2/100) => x = 300/2 => x = 150`
`y = (200/100 - 9)/(-4/100) => y = 700/4 => y = 175`
Hence, evaluating the number of products produced per week to reach the maximum profit yields that for `150` products produced per week, the profit is of `$175.`
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calendarEducator since 2012
write1,284 answers
starTop subjects are Math and Science
The formula of profit is:
Profit = Revenue - Cost
Plug-in Revenue= R(x) and Cost=C(x).
Profit= R(x) - C(x)
Since `R(x) = 6x-1/100x^2` and `C(x)=50+3x` , then,
Profit = `6x-1/100x^2 - (50+3x)`
Profit = `6x-1/100x^2 - 50-3x`
Profit =`-1/100x^2+3x-50`
Notice that the profit function is in quadratic form. So its graph is a parabola.
In parabola, if the coefficient of x^2 is negative, the vertex is the maximum point.
Since the coefficient of x^2 in the profit function is negative, then its vertex is the maximum point.
To determine its vertex (h, k), use the formula,
`h=-b/(2a)`
where a is the coefficient of x^2 and b is the coefficient of x.
`h = -3/(2*(-1/100)) = -3/(-1/50) = -3*-50/1=150`
Then, substitute x with the value of h to the profit function to get k.
`k=Profit = -1/100x^2+3x-50=-1/100*150^2 +3*150-50`
`k=-225 +450-50=175`
Base on this, the vertex is (150, 175).
Hence, the company has to sell 150 products per week to attain the maximum profit of $175 per week.