#### Answer

For maximum profit, the company should manufacture $75$ cabinet per day and $\$1200$ is maximum profit.

#### Work Step by Step

The profit of the company is modelled as $p\left( x \right)=-{{x}^{2}}+150x-4425$ which is a quadratic equation with $a<0$. Thus, it would have a maximum value at
$x=-\frac{b}{2a}$
$\begin{align}
& x=-\frac{150}{2\left( -1 \right)} \\
& x=75
\end{align}$
$p\left( x \right)=-{{x}^{2}}+150x-4425$
Put $x=75$ in the above equation to maximize the company profit,
$\begin{align}
& p\left( 75 \right)=-{{\left( 75 \right)}^{2}}+150\left( 75 \right)-4425 \\
& =-5625+11250-4425 \\
& =-5625+6825 \\
& =1200
\end{align}$