Maximum profit problems... Urgent please help!!!!!?

Question

The green mountain Inn can rent all its 200 rooms when it charges $28 per night for a room, but the manager wants to increase profits. He finds that for each $1 increase in room rate, 4 fewer rooms are rented.
If the cost of cleaning an occupied room is $8 what should the manager charge per night for a room to maximize profits?

Answer 1

Hi,

The Manager should charge $43 per night to earn maximum profit.

Proof:
An increase in $15 would mean a reduction of 15*4=60 rooms; A total of 140 rooms;
Total revenue = 140*$43=$6020
Cleaning charges=140*$8=$1120
Profit earned = 6020-1120=$4900

This scenario is called as Break-even; If you try with $44 or $42 you will understand that the profit comes down to $4896, though there could be a chage in revenue earned.

Hope it is clear.
All the Best!!
Cheers!!

Answer 2

The formula for profit = income - cleaning expense

Let x be increase in the room rate.
then

Profit, p = (200 - 4x)(28 + x) - 8(200-4x)
p = (200 - 4x)(20 + x)
= 4000 + 120x - 4x^2

For maximum profit, do a differentiation:
dp/dx = 120 - 8x = 0
x = 15.

So the manager should increase the room rate to $28+15 = $43,

Then the income is 43 x [200-15(4)] = $6020

Cleaning = 8 x [200-15(4)] =$1120

Profit = 6020 - 1120 = $4900

Answer 3

Now say if he increased the rent by $x then the no. of rooms will be fewer that 4x and the cleaning charges will be 8 * 4x

So. it will be ,

(28 + x) ( 200 - 4x) - 8* 4x > 28* 200 or

5600 + 200x - 4x^2 - 112x - 32x >5600

4x^2 < 56x or x < 14 . So the maximum value of x is $13

Therefore to have maximum profit the rent of the rooms should be $28 + $13 = $ 41

Answer 4

Let hike be of x$

So, now room rent is (28+x) $

Rooms which are rented (200-4x)

Cost of cleaning empty rooms is $8*4x = $ 32x

So, we have to maximize

f(x) = (28+x)(200-4x) - 32x = 4{(28+x)(50-x) - 8x}

f'(x) = 4{(50-x)-(28+x)-8} = 0 (for maximum)

or 50-x - 28-x-8 = 0
or 2x = 14
or x =7

Thus the manager should charge (28+7)$ = $ 35 to maximize profit.

Answer 5

if the room rate is r
profit = (r-8)*(200 - 4(r-28) )
= 200(r-8) - 4( r^2-36r+224)
= -4(r^2-86r+624)
= -4(r-8)(r-78)
profit maximised at (78+8)/2 = 43 per night