Question

A company manufactures and sells x television sets per month. the monthly cost and​ price-demand equations are ​c(x)=75,000 50x and p(x)=300− x 30​, 0≤x≤9000. ​(a) find the maximum revenue.

Answer 1

182.5 is the maximum revenue.

What do revenue means?

• Revenue is the total amount of income generated by the sale of goods or services related to the company's primary operations.
• Revenue, also known as gross sales, is often referred to as the "top line" because it sits at the top of the income statement.
• Income, or net income, is a company's total earnings or profit.

Revenue = p(x) × x = ( 300 - x /30)x

⇒R(x) = 300x - x²/30

R'(x) = $\frac{d}{dx} (300 - \frac{x^{2} }{30} ) = \frac{d}{dx} (300x) - \frac{d}{dx} (\frac{x^{2} }{30})$

    300 - x /15 = 0

⇒ x = 300 . 15 = 4500

∴ Max Revenue = R(4500) = 300 . 4500 - 4500²/30 = 675000 $

Profit = Revenue - cost

P(x) = 300x -  x²/30 - (75000 + 60x)

P'(x) = d/dx ( 300x - x²/30 - ( 75000 + 60x)

       = d/dx (300x) - d/dx (x²/30) - d/dx ( 75000 + 60z))

       = 300 - x /15 -60

       = 240 - x /15

P'(x) = 0 ⇒ x = 240 . 15  = 3600

P(3600) = 300 .3600 - 3600²/30 - (75000 +60 . 3600) =357000

∴ Max profit is 357000$ when 3600 sets are manufactured

After taxation , p(x) = 300x  - x²/30 - ( 75000 + 60x) -5x

P'(x) = 300x  - x²/30 - ( 75000 + 60x) -5x

      = d/dx (300x) - d/dx (x²/30) - d/dx ( 75000 + 60x ) - d/dx (5x)

      = 300 - x/15 - 60 -5 = 235 - x/15

P'(x) = 0 ⇒ x = 235 . 15 = 3525

Pmax = P(3525) = 300 .3525 - 3525²/30 - ( 75000 + 60 .3525) - 5 .3525= 678375/2  (Decimal ; 339187.5)

P(3525) = 300 - 3525/30 = 365/2 = 182.5

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