A polynomial equation has only imaginary roots and no real root. Which options CANNOT be the degree of that polynomial? (Select
2 answers:
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Answer:
a) p(x) = 300 - 3
b) P(x) = -3 x² + 285 x
c) Price of per room per day = $ 157.5
when Number of rooms occupied , x = 47.5
Step-by-step explanation:
Given - A motel owner observes that when a room is priced at $60 per day, all 80 rooms of the motel are occupied. For every $3 rise in the charge per room per day, one more room is vacant. Each occupied room costs an additional $15 per day to maintain.
To find - a) Find the demand function, expressing p, the price charged for each room per day, as a function of x, the number of rooms occupied.
b) Find the profit function P(x).
c) Find the price of per room per day the motel should charge in order to maximize its profit.
Proof -
a)
Let
(x, y) be the point
where x represents number of rooms occupied
and y represents price of room per day.
Now,
Given that,
a room is priced at $60 per day, all 80 rooms of the motel are occupied.
So, point becomes (80, 60)
And given that For every $3 rise in the charge per room per day, one more room is vacant.
So, point becomes (79, 63)
Now, we have two points (80, 60), (79, 63)
Let us assume that,
p(x) be the price charged for each room per day
Now,
By using point - slope formula , we get
p -60 =
⇒p -60 = (-3)(x-80)
⇒p-60 = 240 -3 x
⇒p(x) = 240 + 60 -3 x
⇒p(x) = 300 - 3 x
b)
Given that,
Each occupied room costs an additional $15 per day to maintain.
Let C(x) be the cost function,
Then C(x) =15 x
now,
Revenue function,
R(x) =x*p
= x*(300 -3 x )
= 300 x - 3 x²
⇒R(x) = 300 x - 3 x²
Now,
We know
Profit function = Revenue function - Cost function
⇒P(x) = R(x)-C(x)
⇒P(x) = (300 x -3 x²) -15 x
⇒P(x) = -3 x² + 285 x
c)
P'(x) = -6 x +285
For Maximize profit , Put P'(x) = 0
⇒-6 x+ 285 =0
⇒6 x= 285
⇒x =
⇒x= 47.5
∴ we get
Maximize profit is when price, p = 300 - 3x
= 300 -3(47.5)
= $157.5
⇒Price of per room per day = $ 157.5
when Number of rooms occupied , x = 47.5