Question

A hotel has 210 units. All rooms are occupied when the hotel charges $90 per day for a room. For every increase of x dollars in the daily room rate, there are x rooms vacant. Each occupied room costs $24 per day to service and maintain. What should the hotel charge per day in order to maximize daily profit

Answer 1

Answer:  $\ 162$

Step-by-step explanation:

Given

There are 210 units of hotel

Each room charge $90 when fully occupied

occupied room charges is $24 per day

Suppose x rooms are vacant, so, there 210-x occupied

Daily expense on maintenance$=(210-x)\times 24$

Earning from 210-x rooms$=(210-x)(90+x)$

Profit(P)=Earning - Expense

$\Rightarrow P=(210-x)(90+x)-(210-x)24\\\Rightarrow P=21,600+210x-90x-x^2-5040+24x\\\Rightarrow P=-x^2+144x+16,560$

To get the maximum profit, determine the derivative of P w.r.t. x and equate it to zero

$\Rightarrow \dfrac{dP}{dx}=-2x+144\\\\\Rightarrow -2x+144=0\\\\\Rightarrow x=72$

Charges for maximum profit is $90+72=\ 162$

Number of occupied rooms are $210-72=138$

Answer 2

The hotel charge per day will be $162  in order to maximize daily profit.

Given that,  Hotel has 210 units

Let us consider that x rooms are vacant.

Total occupied rooms = $(210-x)$

Since, Each occupied room costs $24 per day to service and maintain

So, Total expense = $24*(210-x)=5040-24x$

Since, For  all rooms are occupied when the hotel charges $90 per day for a room. For every increase of x dollars in the daily room rate, there are x rooms vacant.

So, Total earing = $(210-x)*(90+x)=-x^{2} +120x+21600$

Profit = Total earning - Total expense

$P=-x^{2} +120x+21600-5040+24x\\\\P=-x^{2} +144x+16560$

In order to maximize profit, Differentiate profit expression with respect to x.

    $\frac{dP}{dx}=\frac{d}{dx}(-x^{2} +144x+16560) \\\\\frac{dP}{dx}=-2x+144=0\\\\\\x=\frac{144}{2}=72$

Thus, Hotel charges per day for maximum profit = $90+x=90+72=162$

Learn more:

https://brainly.com/question/19578119