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The recruit function is given as p(x) = 41 − 24x − 18x2 Therefore, p’(x) = - 24 - 36x P’’(x) = -36 Now, p’(x) = 0 => x = -24/36 = -2/3 Also, P’’(-2/3) = -36 < 0 By second derivative test, x = -2/3 is the point of local maxima of p. Therefore, Maximum profit = p(-2/3) => 41 – 24(-2/3) – 18(-2/3)2  => 41 + 16 – 8 = 49 Hence, the maximum recruit that the company can make is 49 people.
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