DavidP

11-03-2003, 01:24 AM

I am having trouble on a math problem. I have tried everything I know to do, but I cannot seem to find the answer. (Well, I get an answer, but I do not think it is correct).

I will write out the problem, show you my work, and my answer. Tell me your thoughts.

The Problem:

The cost per unit for the production of a certain radio model is $60. The manufacturer charges $90 per unit for orders of 100 or less. To encourage large orders, the manufacturer reduces the charge by $0.15 per radio for each unit ordered in excess of 100.

a. Write the profit as a function of x.

b. How many radios should he sell to maximize profit?

My work:

So there are a few different functions here.

The cost to produce a radio is $60. Therefore:

c ( x ) = 60 x

The revenue the guy gets from selling radios is dependent on how many radios he sells. If it is less than 100 radios, it is $90 per radio, otherwise it is $90 per radio minus .15 cents per every radio over 100.

Therefore:

f ( x ) =

if x <= 100

then return 90x

else if x > 100

then return 90x - .15(x - 100)

Profit is revenue minus cost. Therefore:

p(x) = f(x) - c(x)

Therefore, if we are under 100 radios, p(x) is this:

90x - 60x

If we are above 100 radios:

(90x - .15(x-100)) - 60x

Simplify:

(90x - .15x + 15) - 60x

(89.85x + 15) - 60x

p(x) = 29.85x + 15

This function is completely linear. It has no max and no min. Profit never maximizes. It approaches infinity.

Am I doing something wrong?

I will write out the problem, show you my work, and my answer. Tell me your thoughts.

The Problem:

The cost per unit for the production of a certain radio model is $60. The manufacturer charges $90 per unit for orders of 100 or less. To encourage large orders, the manufacturer reduces the charge by $0.15 per radio for each unit ordered in excess of 100.

a. Write the profit as a function of x.

b. How many radios should he sell to maximize profit?

My work:

So there are a few different functions here.

The cost to produce a radio is $60. Therefore:

c ( x ) = 60 x

The revenue the guy gets from selling radios is dependent on how many radios he sells. If it is less than 100 radios, it is $90 per radio, otherwise it is $90 per radio minus .15 cents per every radio over 100.

Therefore:

f ( x ) =

if x <= 100

then return 90x

else if x > 100

then return 90x - .15(x - 100)

Profit is revenue minus cost. Therefore:

p(x) = f(x) - c(x)

Therefore, if we are under 100 radios, p(x) is this:

90x - 60x

If we are above 100 radios:

(90x - .15(x-100)) - 60x

Simplify:

(90x - .15x + 15) - 60x

(89.85x + 15) - 60x

p(x) = 29.85x + 15

This function is completely linear. It has no max and no min. Profit never maximizes. It approaches infinity.

Am I doing something wrong?