Answer to Question #114339 in Financial Math for sana

Question #114339
a bakery finds that the price they can sell cakes is given by the function p=580-10k where x is number of the cakes sold per day, and p is price. the total cost function of the company is given by c= ( 30 + 5x)^2 where x is previously defined and c is the total cost. A. find the revenue and the marginal revenue functions [ Hint: revenue is the price multiplied by quantity i.e. revenue = price* quantity]
B. find the fixed cost and marginal cost function [ Hint: fixed cost does not change with quantity produced]
C. find the profit function [ Hint: profit is revenue minus total cost]
D. find the quantity that maximums profit.
1
Expert's answer
2020-05-08T18:35:00-0400

A. find the revenue and the marginal revenue functions

The total revenue function is computed using the following formula:


"\\text{Revenue Function}=\\text{Price}\u00d7\\text{Quantity}"

"\\text{Total revenue}=[580\u221210x]\u2217x"

"\\text{Total revenue TR}=580x\u221210x^2"

The marginal revenue is the derivative of the total revenue function

"\\text{Marginal revenue}=\\dfrac{\\Delta TR}{\\Delta x}"


"\\text{Marginal revenue}=580\u221220x"

B. find the fixed cost and marginal cost function

The total cost function is computed as follows:

"\\text{Total cost function TC}=(30+5x)(30+5x)"


"\\text{Total cost function}=900+300x+25x^2"

The fixed cost based on the above cost function is 900 since it is not influenced by the level of output.

The marginal cost function is the derivative of the total cost function.

"\\text{Marginal cost}=\\dfrac{\\Delta TC}{\\Delta x}"


"\\text{Marginal cost}=300+50x"

C. find the profit function

"\\text{Profit function=Revenue function-Cost function}"

"\\text{Profit function}=580x\u221210x^2\u2212[900+300x+25x^2]"

"\\text{Profit function PF} =\u221235x^2+280x\u2212900"

D. find the quantity that maximums profit.

The quantity that maximizes profit will be computed by finding the first order condition of the profit function and equate it to 0 and solve for X.

"\\dfrac{\u0394PF}{\u0394x}\u200b=\u221270x+280"


"\u221270x+280=0"


"\u221270x=\u2212280"


"x=\\dfrac{\u2212280}{\u221270\u200b}"


"x=4"

The revenue maximizing quantity is 4 units.


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