Find the maximum profit that a company can make, if the profit function is given by
$ p(x) = 41 - 24x - 18x^2 $

Q: Find the maximum profit that a company can make, if the profit function is given by $ p(x) = 41 - 24x - 18x^2 $

Explanation: We have: \[ p(x) = 41 - 24x - 18x^2 \Rightarrow p'(x) = -24 - 36x \Rightarrow p''(x) = -36 < 0 \] Set derivative to zero: \[ p'(x) = 0 \Rightarrow -24 - 36x = 0 \Rightarrow x = -\frac{2}{3} \] By second derivative test, local maximum occurs at $ x = -\frac{2}{3} $ Now, compute: \[ p\left( -\frac{2}{3} \right) = 41 - 24\left( -\frac{2}{3} \right) - 18\left( -\frac{2}{3} \right)^2 = 41 + 16 - 8 = 49 \] $\textbf{Maximum profit}$ = 49

Question

Find the maximum profit that a company can make, if the profit function is given by  
$ p(x) = 41 - 24x - 18x^2 $

Accepted Answer

Explanation: We have:  
$$
p(x) = 41 - 24x - 18x^2
\Rightarrow p'(x) = -24 - 36x
\Rightarrow p''(x) = -36 < 0
$$

Set derivative to zero:
$$
p'(x) = 0 \Rightarrow -24 - 36x = 0 \Rightarrow x = -\frac{2}{3}
$$

By second derivative test, local maximum occurs at $ x = -\frac{2}{3} $

Now, compute:
$$
p\left( -\frac{2}{3} \right) = 41 - 24\left( -\frac{2}{3} \right) - 18\left( -\frac{2}{3} \right)^2
= 41 + 16 - 8 = 49
$$
$\textbf{Maximum profit}$ = 49

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