## Precalculus (6th Edition) Blitzer

The amount the hotel should charge for each room to maximize its daily profit is $\95$ and the maximum daily profit is$\21,675$.
Let the amount the hotel should charge for each room to maximize its daily profit be x. The number of rooms rented depends on the rent, where the number of rooms is the original number $300$, minus the number lost to rent increase. Thus, the number of rooms rented are $300-3x$ Let the daily profit of a hotel be P The rent is $80+x$. And the cost of service per day is cost multiplied by the number of rooms rented. The cost of service per day is $10\left( 300-3x \right)$. Recall that the daily profit of a hotel is the product of the number of rooms and the rent minus the cost of service per day. Thus, \begin{align} & P\left( x \right)=\left( 300-3x \right)\left( 80+x \right)-10\left( 300-3x \right) \\ & =24000+60x-3{{x}^{2}}-30x-3000 \\ & =-3{{x}^{2}}+90x+21000 \end{align} Maximize $P\left( x \right)=-3{{x}^{2}}+90x+21000$. Compare it with the standard form $f\left( x \right)=a{{x}^{2}}+bx+c$ The value of $a$ is $-3$ , $b$ is $90$ and $c$ is $21000$ Since, $a<0$ , the function has maximum value at $x=\frac{-b}{2a}$. \begin{align} & x=\frac{-b}{2a} \\ & =\frac{-90}{2\times \left( -3 \right)} \\ & =\frac{-90}{-6} \\ & =15 \end{align} Recall that the rent is $80+x$. Substitute $15$ for x in $80+x$ , \begin{align} & 80+x=80+15 \\ & =95 \end{align} Substitute $15$ for x in $P\left( \frac{-b}{2a} \right)$ . \begin{align} & P\left( 15 \right)=-3{{x}^{2}}+90x+21000 \\ & =-3{{\left( 15 \right)}^{2}}+90\left( 15 \right)+21000 \\ & =21675 \end{align} Hence, the amount the hotel should charge for each room to maximize its daily profit is $\$95$and the maximum daily profit is$\$21,675$.