The maximum profit that a company can make if the profit function is…

$P(x)=41+24 x-18 x^2$

differentiating w.r.t x and setting it equal to zero to find critical points

$P'(x)=\frac{d}{d x}\left(41+24 x-18 x^2\right)$

$P'(x)=24-36 x=0$

$\Rightarrow 24=36 x$

$x=24 / 36=2 / 3$

differentiating it again 

$P''(x)=\frac{d}{d x}(24-36 x)=-36$

for $x = \frac{2}{3}$

P''(x) < 0

$x = \frac{2}{3}$ → point of maxima

so maximum value of P(x)

$P(\frac{2}{3}) = 41 + 24 × \frac{2}{3}-18×\frac{2×2}{3×3}$

= 41 + 16 - 8

= 41 + 8

= 49