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Answer :
To maximize profit, Mr. Graham should charge $70 per night. This is found by creating a profit function $P(x) = (80-x)(60+x) - 4(80-x)$, where $x$ is the increase in room rate, and maximizing it using either algebraic methods (finding the vertex of the parabola) or calculus methods (setting the derivative to zero). The assumptions made are a linear relationship between price increase and occupancy decrease, fixed maintenance costs, and no other revenues or costs.
Profit Maximization for Mr. Graham's Hotel
Mr. Graham needs to determine the optimal nightly rate for his hotel to maximize profit, considering that for every $1 increase in room rate, one less room will be rented. Assuming all 80 rooms are rented at $60 each, as the baseline, each $1 increase will result in losing one occupancy. So, if the room rate is increased by $x, there will be $(80-x)$ rooms rented at $(60+x)$ dollars.
The total revenue is $R(x) = (80-x)(60+x)$, and the cost for the occupied rooms is $C(x) = 4(80-x)$, since unoccupied rooms do not incur maintenance costs. The profit function would then be $P(x) = R(x) - C(x) = (80-x)(60+x) - 4(80-x)$.
To maximize profit using algebraic techniques, we can expand the profit function, $P(x) = 4800 + 20x - x^2$, and calculate the vertex of the parabola (which represents maximum profit), where $x = -b/(2a)$ from the general form $ax^2 + bx + c$.
In this case, $a=-1$ and $b=20$, so the optimal increase in rate is $x = -20/(2(-1)) = 10$. Hence, the maximum profit will be when the room rate is $60 + 10 = $70 per night.
For calculus methods, we take the derivative of the profit function, $P'(x) = d/dx(4800 + 20x - x^2) = 20 - 2x$, and set it equal to zero to find the critical points. Solving $P'(x) = 0$ gives $x = 10$, confirming the algebraic result. Substituting $x=10$ into the profit function gives us the maximum profit.
Assumptions:
- The relationship between room rate increase and occupancy decrease is linear and consistent.
- All rooms are equivalent and maintenance costs are fixed.
- There are no other costs or sources of revenue.
Conclusion
Mr. Graham should set the room rate at $70 per night to maximize his profit. The corresponding maximum daily profit would be $P(10) = (80-10)(60+10) - 4(80-10) = $4620
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