Given
$P\left(x\right)=190-0.5x\phantom{\rule{0ex}{0ex}}C\left(x\right)=2500+0.75{x}^{2}$
The total Revenue is
$\begin{array}{rcl}R\left(x\right)& =& P\left(x\right)x\\ & =& \left[190-0.5x\right]x\\ & =& 190x-0.5{x}^{2}\end{array}$
The total profit is
$\begin{array}{rcl}P& =& R-C\\ & =& 190x-0.5{x}^{2}-\left[2500+0.75{x}^{2}\right]\\ & =& 190x-0.5{x}^{2}-2500-0.75{x}^{2}\\ & =& -1.25{x}^{2}+190x-2500\end{array}$
To find number of suits must the company produce and sell in order to maximize profit differentiate $P=-1.25{x}^{2}+190x-2500$ with respect to x we get
$\begin{array}{rcl}P\text{'}& =& -1.25\left(2x\right)+190-0\\ & =& -2.50x+190\end{array}$
For finding maximum number of suits
$P\text{'}=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2.50x+190=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2.50x=190\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{190}{2.50}\phantom{\rule{0ex}{0ex}}\Rightarrow x=76$
So, 76 suits must the company produce and sell in order to maximize profit
Maximum profit is
$\begin{array}{rcl}P\left(76\right)& =& -1.25{\left(76\right)}^{2}+190\left(76\right)-2500\\ & =& -1.25\left(5776\right)+14,440-2500\\ & =& -7220+11,940\\ & =& 4720\end{array}$
To find price per suit must be charged in order to maximize profit
$\begin{array}{rcl}p& =& 190-0.5\left(76\right)\\ & =& 190-38\\ & =& 152\end{array}$