Demand for rooms, of a hotel which has #58# rooms, is a function of price charged given by #u(p)=p^2-12p+45#. Find out at what price the revenue is maximized and what is the revenue?

Question

Emma Aldridge · Accepted Answer

There is no limit to revenue for #p>13#, but demand cannot be fulfilled beyond #58# rooms, which is for #p=13# and maximum revenue at this level is #754#.

As the number of rooms that will be occupied, based on the price being charged, is #u(p) = p^2 - 12p + 45# with #u(p)<=58# and #u(p)inI#. As such revenue #r# will be given by #p(p^2-12p+45)# and this will be maximized when #d/(dp)r(p)=0#, where #r(p)=p^3-12p^2+45p# and second derivative #d^2/(dp)^2r(p)<0# for maxima and #d^2/(dp)^2r(p)>0# for minima. As #d/(dp)r(p)=3p^2-24p+45# and #3p^2-24p+45=0# and dividing each term by #3#, we get #p^2-8p+15=0# i.e. #(p-5)(p-3)=0# and as #d^2/(dp)^2r(p)=6p-24=6(p-4)# while for #p=5#, #d^2/(dp)^2r(p)=6# for #p=3#, #d^2/(dp)^2r(p)=-6# Hence, we have a local maxima at #p=3# and a local minima at #x=5# At #p=3#, we have #r(3)=3^3-12xx3^2+45xx3=54# and at #p=5#, we have #u(5)=5^3-12xx5^2+45xx5=50#, but the latter is a local maxima and as subsequently #r(p)# continues to rise and is limited only by #u(p)<=58#. And at #u(p)=58# and #p^2-12p+45=58# is #p^2-12p-13=0# i.e. #(p-13)(p+1)=0# Revenue is maximized at #p=13#, where it is #13^3-12xx13^2+45xx13=2197-2028+585=754#, when occupancy is #58#. However, as even beyond #p=13# i.e. for #p>13#, demand for rooms continues to increase. Hence, answer is that there is no limit post the price #p>13#, graph{x^3-12x^2+45x [-5, 15, -50, 800]}

Evelyn Agard · Answer

undefined To find the price at which revenue is maximized, we need to find the critical points of the revenue function. The revenue function, R(p), is given by the product of price and demand, which is u(p) * p. 
R(p) = p(u(p)) = p(p^2 - 12p + 45). 
To find critical points, we differentiate R(p) with respect to p and set it equal to zero, then solve for p.
R'(p) = 3p^2 - 24p + 45.
Setting R'(p) = 0 and solving for p, we get:
3p^2 - 24p + 45 = 0.
Solving this quadratic equation yields p = 3 or p = 5. 
To determine which value of p maximizes revenue, we use the second derivative test. 
R''(p) = 6p - 24. 
For p = 3, R''(3) = 6(3) - 24 = 18 - 24 = -6, indicating a maximum. 
For p = 5, R''(5) = 6(5) - 24 = 30 - 24 = 6, indicating a minimum. 
Therefore, revenue is maximized when p = 3. 
To find the maximum revenue, substitute p = 3 into the revenue function:
R(3) = 3(3^2 - 12(3) + 45) = 3(9 - 36 + 45) = 3(18) = 54. 
So, the price at which revenue is maximized is $3, and the maximum revenue is $54.