Calculus: Finding revenue using derivatives.?

Question

A manufacturer estimates that D(p) = 5000e^-0.02 units of a particular commodity will be sold when the price is p dollars per unit. Determine the market price p that will result int the largest revenue.

I am not finding for the maximum D(p), I am finding the maximum R(p), which is [D(p)]p. What I'm trying to look for is the price which will result to my highest revenue, not my highest D(p) or demand.

Answer 1

The function D gives only the amount of units sold. The revenue is the product of price and number of units sold:

R(p) = p * D(p)= p*exp(-0.02p)

This function has a maximum:

dR(p)/dp=0

exp(-0.02*p)*(1-0.02*p)=0

because exp(-0.02)>0 for any p

1-0.02*p=0 <=> p=50

Proofing the maximum with the second derivative

d²R/dp²=-0.04*exp(-0.02*p) +0.0004*p*exp(-0.02*p)
=0.04*exp(-0.02*p)(p-100)
which is for p=50
= -200*exp(- 100) < 0 => maximum

Conclusion: the manufacturer will gain a maximal revenue, if he sells for a price of 50$.

Answer 2

Since the right side of the function does not include p, I will assume there is a typo, and the actual function is

D(p) = 5000 * e^(-0.02 * p)

Note that p >= 0, since it is a price. Since D(p) is a decreasing function (plot some values of p to see it, and check that D'(p) is always < 0), D(p) is maximum at p = 0, as common sense would dictate.

If I misinterpreted the definition of D, and it isn't always decreasing, here is a general explanation:

The value of p such that D(p) is an extreme (maximum or minimum) is given by D'(p) = 0. Solve this equation for p. Plot D(p) to know if the solutions are maxima or minima.

You will need the chain rule to differentiate D(p). The general case is:

(f(g(x))' = f'(g(x)) * g'(x)

Remember that e^x is a function, too: exp(x), which is its own derivative.

Answer 3

D(p) = 5000e^0.02p ( I assume )
if price is p then D is sold

max revenue is when R(p) = p*D(p) is maximal
R'(p) = D(p) - 0.02p * D(p)
R'(p) = 0 when 1-0.02p = 0 thus for p = 1/0.02 = 50

so when the price is 50 ( you still have to check th/at this is a maximum of R(p) and not a minimum I leave that to you )

Answer 4

I have taken the liberty of inserting a "p" into your exponent.

Nevertheless, this can't be right: D is a maximum at P=0.

Answer 5

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