Algebra

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Suppose that the price per unit in dollars of a cell phone production is modeled by
p = $55 − 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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  1. 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 = -b%2F%282a%29 ( referring to the general form of a quadratic function q(x) = ax%5E2+%2B+bx+%2B+c ),

    which at given conditions is x = -35%2F%282%2A%28-0.0125%29%29 = 35%2F0.025 = 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%5Bmax%5D = R(1400) = -0.0125%2A1400%5E2+%2B+35%2A1400 = 24500 thousands of dollars.

    Answer. The maximum revenue is 24500 thousands of dollars achieved at the production level of 1400 thousand of phone units .

    Hope this help you with your homework :3

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