# demand function

There are two different problems of calculus (see the attachment).

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Please find solutions/explanations attached herewith.

1) Consider the demand function for selling x widgets given by

p(x) = 30(ln(40)+1) -30ln(x)

a) Find the revenue generated by selling x widgets.

Revenue R(x) = xp(x)

R(x) = x*(30(ln(40)+1) -30ln(x))

R(x) = 30x(ln(40)+1) -30xln(x)

b) Assume that the production level is related to the number of employees by the formula x =10n where n is the number of employees. Find the revenue generated as a function of n employees.

Put x = 10n

R(n) = 30*(10n) (ln(40)+1) -30*10n*ln(10n)

R(n)= 300n(ln(40)+1) - 300nln(10n)

R(n)= 300n(ln(40)+1) -300n(ln(10)+ln(n))

R(n)= 300nln(40) + 300n - 300n(1n(10) + ln(n))

R(n)= 300nln(40) + 300n - 300nln(10) - 300nln(n)

R(n)= 300nln(40) - 300nln(10) + 300n - 300nln(n)

R(n)= 300nln(40/10) + 300n - 300nln(n)

R(n)= 300nln(4) + 300n - 300nln(n)

c) Find all local extrema of the function R(n) that you found in part (b).

R'(n) = 300ln(4) + 300 - 300ln(n) -300n*(1/n)

R'(n) = 300ln(4) + 300 - 300ln(n) -300

R'(n) = 300ln(4) - 300ln(n)

Put R'(n) = 0

300ln(4) - 300ln(n) = 0

300ln(4) = 300ln(n)

=> n = 4

n = 4 is the critical point.

Let us take two points around n = 4, say n = 3 and n = 5

R'(3) = 300ln(4) - 300ln(3) = 300ln(4/3) = 86.30 > 0

R'(5) = 300ln(4) - 300ln(5) = 300ln(4/5) = -66.94 < 0

Function is increasing when n = 3 and decreasing when n = 5. Thus, n = 4 is the maximum point.

Revenue at n = 4 = 300*4*ln(4) + 300*4 - 300*4ln(4) = $1200

Local maximum of R(n) is (4, 1200)

d) How many employees should be hired to maximize the revenue? What is this maximum revenue?

Number of employees hired x = 10n = 10*4 = 40

Maximum revenue = 300*4*ln(4) + 300*4 - 300*4ln(4) = $1200

2) A company's cost and revenue functions are given by the equations

a) Find the equation for the profit.

Profit P(x) = R(x) - C(x)

b) Use the critical values of the profit function to determine (1) the production levels for which the profit is increasing, (2) the production levels for which the

profit is decreasing, and (3) the production level for which the profit is maximized.

Let us find P'(x)

Put 0

Multiply by -100

Divide by 25

â‡¨ x = -7.70977, 10.3764

Ignore the negative value as function is only defined in the interval .

Let us take some test values say x = 9, 11

Thus, profit function is increasing in the interval [0, 10.3764] and decreasing in the interval [10.3764, 20].

Production level will be maximum at x = 10.3764.

Maximum profit =

(rounded to 2 decimal places)

c) In economics, people generally find the production level that corresponds to the maximum profit by solving equation MR(x) = MC(x). Explain why this makes sense knowing what we know about critical values, local extrema, marginal revenue and marginal cost.

Profit is maximized at the point at which an additional increment to output leaves profit unchanged.

P = R - C

P' = R' - C'

P'= 0

â‡¨ R' - C' = 0

â‡¨ MR = MC

Marginal revenue is the slope of revenue curve and marginal cost is the slope of total cost curve.

Or we can say profit is maximized at the point where slopes of the revenue and cost curves are equal.

That is R' = C'

=> MR = MC.

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