Expert: Scotto - 11/20/2010

Question
QUESTION: I am sorry for the last message i sent professor, these are the revised questions.

Determine the location of each local extremum of the function.
1) f(x)=-x^(3)-9x2-24x-2

Solve the problem.
2) If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in a certain city, where p(x) =71-(x)/(24) . How many candy bars must be sold to maximize revenue?

3.Find the number of units that must be produced and sold in order to yield the maximum profit, given the following
equations for revenue and cost
R(x) = 40x -0.5x^(2)
C(x) = 7x + 10.

4. Suppose c(x)= x^(3)-24x^(2)+20,000x  is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.

5. Suppose a business can sell x gadgets for p = 250 - 0.01x dollars apiece, and it costs the business c(x)= 1000+25x dollars to produce the x gadgets. Determine the production level and cost per gadget required to maximize profit.

6) Find the number of units that must be produced and sold in order to yield the maximum profit, given the following
equations for revenue and cost
R(x) = 5x
C(x) = 0.001x^(2) + 1.3x + 70.



ANSWER: Find extreme points of f(x) = -x^3 - 9x^2 - 24x - 2
To find the extremum, find the derivative and set it equal to 0.
It looks like the derivative is f'(x) = -3x^2 - 18x - 24.
Set this equal to 0, divide by -3, and then factor.
This will show where the zeros are of the derivative.
Where the zeros are is where the extema are.

2) x is in thousands; price is p(x) = 71-(x)/(24); find the maximum.
Revenue r(x) = xp(x) = 71x - x^2/24.
From here, r'(x) = 71 - x/12; set r'(x) = 0; find x.

3. Given revenue R(x) = 40x -0.5x^(2) and cost C(x) = 7x + 10;
maximize revenue; determine cost
Find R'(x), set to 0, solve for x, substiture that x into C(x)

4. Cost c(x) = x^3 - 24x^2 + 20,000x; minimize cost
From here, c'(x) = 3x^2 - 48x + 20,000.
This has no zeros; minimum cost is at 0 production;
are all the signs correct?

5. Price p(x) = 250 - 0.01x is price on each;
there are x gadgets, so total revenue r(x) = xp(x);
cost is c(x)= 1000+25x; profit f(x) = r(x) - c(x);
f(x) = 250x - 0.01x^2 - 1000 - 25x;
Find f'(x), set f'(x) = 0; solve for x;
put back into p(x) to find profit.

6) Revenue is R(x) = 5x and cost is C(x) = 0.001x^(2) + 1.3x + 70.

It is known that profit is revenue minus cost, so P(x) = R(x)-C(x).
Put in R(x) and C(x) to find P(x); the x terms will combine.

To solve, find P'(x), set it to 0, solve for x,
and this is the number of units.


---------- FOLLOW-UP ----------

QUESTION: I'm sorry for the bother again professor. You have been a great help,but on the last questions you gave me steps to get the answer but no answer, could you send me the answers for i could see if i am following the steps correctly. thank you.

along with the answer for this question that you gave me steps for...

7.- The demand function for a product is given by P= 20 - 0.02X, 0<x<1000.
a) Find the price elasticity of demand when x=560
I believe the elasticity is the derivative, and that is -0.02.

b) Find the values of x and p that maximize the total Revenue
If the total profit is (20 - 0.02x)x = 20x - 0.02x^2, then to maximize it, set the derivative equal to 0.

c) For the value of X that maximizes the Revenue , show that the elasticity is UNIT ELASTICITY.

To find the value of X that maximizes the revenue, take the derivative of the funciton and set it to 0, then solve for x.
UNIT elasticity means that at this point, an increase of a fixed amount will decrease production by the same amount.

Answer
a) I believe the elasticity is given by the change in quantity over the change in price.

b) Yes, to maximize it, set the derivative equal to 0.
Once x is known, this can be put back in the equation to maximize it.

c) Once the maximum has been found, put it into the change in quatity over the change in price to see that it is 1.

What you said at the bottom sounds right to me.