I have calculus extra credit but need someone to compare to and check mine with.?

1) The cost is C(x)= (x^3) - (6x^2) + 13x + 15, find the minimum marginal cost.

2) The revenue is R(x)= 200 - [1600/ (x+8) ] - x, find the maximum marginal revenue

3) The average ticket price for a concert at the opera house was $50. The average attendance was 4000. When the ticket price was raised to $52, attendance declined to an average of 3800 persons per performance. What should the ticket price be in order to maximize revenue for the opera house?

4) An artist is planning to sell signed prints of her latest work. If 50 prints are offered for sale, she can charge $400 each. However, if she makes more than 50 prints, she must lower the price of all prints by $5 for each print in excess of 50. How many prints should the artist make in order to maximize revenue?

I have calculus extra credit but need someone to compare to and check mine with.?palace theatre

1) C(x)= (x^3) - (6x^2) + 13x + 15

Marginal Cost is C'(x)

C'(x) = 3x^2 -12x + 13

Minimize C'(x)

0 = 3x^2 -12x + 13

You can solve x yourself.

2) R(x)= 200 - [1600/ (x+8) ] - x

Marginal Revenue is R'(x)

R'(x) = 1600/(x+8)^2 - 1

Maximize R'(x)

0 = 1600/(x+8)^2 - 1

1 = 1600/(x+8)^2

1600 = (x+8)^2

you can solve x yourself.

3) use a table and record the revenue

you will find $45 is the price to maximize the revenue.

At $50 with 4000 attendance, revenue = 200,000

At $52 with 3800 attendance, revenue = 197,600

You must reduce the price to get more than 200,000.

From the data, decrease $2 will increase attendance

by 200

At $48 with 4200 attendance, revenue = 201,600

At $46 with 4400 attendance, revenue = 202,400

At $44 with 4600 attendance, revenue = 202,400

At $42 with 4800 attendance, revenue = 201,600

So maxi revenue occurs at $45.

4)Let x be the # of prints

Revenue R(x) = x[400-5(x-50)]

R(x) = x[400-5x+250]

R(x) = 650x - 5x^2

R'(x) = 650 - 10 x

0 = 650 - 10 x

x = 65

Her revenue = 65* [400-5(15)]=$21,125