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nordsb [41]
3 years ago
13

Hercules Films is deciding on the price of the video release of its film Son of Frankenstein. Its marketing people estimate that

at a price of p dollars, it can sell a total of q = 210,000 − 15,000p copies. (a) What is the revenue function? R(p)= (b) What price will bring in the greatest revenue? p = dollars Second derivative test: Your answer above is a critical point for the revenue function. To show it is a maximum, calculate the second derivative of the revenue function. R"(p)= Evaluate R"(p) at your critical point. The result is , which means that the revenue is at the critical point, and the critical point is a maximum.
Business
1 answer:
natta225 [31]3 years ago
7 0

Answer:

Revenue =  - 15,000p² + 210,000p

Vertex at price = 7

Revenue = 735,000

Explanation:

Revenue = price x quantity

R(p) = (210,000 - 15,000p) = 210,000p - 15,000p²

We calcualte the vertex:

-b/2a = -(-210,000) / (15,000 x 2 ) = 210,000 / 30,000 = 7

-15,000 x 49 + 210,000 x 7 = 735000

R(p) =  - 15,000p² + 210,000p

We derivate using the following identity:

ax^{b} = bax^{b-1}

R(p)' =  - 30,000p + 210,000

R(p)'' =  -30,000

As the second derivate is constant negative there is only one critical point and, is a maximum.

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Some estimates have been presented to Charlene, the Director of Operations (DO) at Holly Farms, which is considering alternative
Marina CMI [18]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

First, we need to sort out the data given properly, so that we can answer to the best.

Data Given:

For Chamber D103:

Installed Cost = -400,000 USD

Annual Operating Cost Per Year = -4000 USD

Salvage Value 10% of P = 40,000 USD

Life = 3 years

Similarly,

For Chamber 490G:

For Chamber D103:

Installed Cost = -250,000 USD

Annual Operating Cost Per Year = -3000 USD

Salvage Value 10% of P = 25,000 USD

Life = 2 years

a.

For Chamber D103

AW Chamber D103 = -400,000 x (A/P, 10%, 3) + 40000 x (A/F,10%, 3) - 4000

So,

(A/P, 10%, 3) = 0.40211    (from the compound interest table )

(A/F,10%, 3) = 0.30211

AW Chamber D103 = -400,000 x 0.40211 + 40000 x 0.30211 - 4000

AW Chamber D103 = -152,760 USD

For Chamber 490G:

AW Chamber 490G = -250,000 x (A/P,10%,2) + 25000 x (A/F,10%, 2) - 3000

So,

(A/P,10%,2) = 0.5762

(A/F,10%, 2) = 0.4762

AW Chamber 490G = -250,000 x 0.5762 + 25000 x 0.4762 - 3000

AW Chamber 490G = -135,143 USD

So, after evaluating both the chambers using the AW method, more economical is the chamber 490G.

b.

Now, we need to change the values to check whether the chamber selection can be changed or not:

so, New values for Chamber D103

P = -300,000 and Salvage value of 30,000

For Chamber D103

AW Chamber D103 = -300,000 x (A/P, 10%, 3) + 30000 x (A/F,10%, 3) - 4000

So,

(A/P, 10%, 3) = 0.40211    (from the compound interest table )

(A/F,10%, 3) = 0.30211

AW Chamber D103 = -300,000 x 0.40211 + 30,000 x 0.30211 - 4000

AW Chamber D103 = -115,570 USD

New Values for Chamber D103

P = -500,000 and the Salvage Value = 50,000

AW Chamber D103 = -300,000 x (A/P, 10%, 3) + 30000 x (A/F,10%, 3) - 4000

So,

(A/P, 10%, 3) = 0.40211    (from the compound interest table )

(A/F,10%, 3) = 0.30211

AW Chamber D103 = -500,000 x 0.40211 + 50,000 x 0.30211 - 4000

AW Chamber D103 = -189,950 USD

Hence,

At P = 300,000 will definitely change the selection to D103 Chamber.

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