SOLUTION: Suppose that the price per unit in dollars of a cell phone production is modeled by p = $35 − 0.0125x,
where x is in thousands of phones produced, and the revenue represente
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-> SOLUTION: Suppose that the price per unit in dollars of a cell phone production is modeled by p = $35 − 0.0125x,
where x is in thousands of phones produced, and the revenue represente
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Question 1071427: Suppose that the price per unit in dollars of a cell phone production is modeled by p = $35 − 0.0125x,
where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x · p.
Find the production level that will maximize revenue. Answer by ikleyn(52772) (Show Source):
You can put this solution on YOUR website! .
Suppose that the price per unit in dollars of a cell phone production is modeled by p = $35 − 0.0125x,
where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x · p.
Find the production level that will maximize revenue.
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According to the condition, the revenue (measured in thousands of dollars) is
R(x) = x*(35-0.0125x),
where x is the number of the phones measured in thousand of units.
So, you need to find the maximum of this quadratic function
R(x) = -0.0125x^2 + 35x.
The maximum is achieved at x = ( referring to the general form of a quadratic function q(x) = ),
which at given conditions is x = = = 1400.
So, the maximum is achieved at the production level 1400 thousand of phone units .
The maximum revenue is the value R(x) at this value of x:
= R(1400) = = 24500 thousands of dollars.
Answer. The maximum revenue is 24500 thousands of dollars achieved at the production level of 1400 thousand of phone units .
