7) If total revenue for a firm is given by $$ R(x)=8000 x-40 x^{2}-x^{3} $$ where $$ x $$ is the number of units sold, find the number of units that must be sold to maximize revenue. Find the maximum revenue.

Question

UpStudy AI · Accepted Answer

To maximize revenue, start by finding the derivative of the revenue function with respect to x:

R(x) = 8000x - 40x² - x³

Take the derivative:

R'(x) = d/dx [8000x - 40x² - x³] = 8000 - 80x - 3x²

Set R'(x) equal to zero to find the critical points:

8000 - 80x - 3x² = 0

Rearrange the equation:

3x² + 80x - 8000 = 0

Solve using the quadratic formula, x = [ -b ± √(b² - 4ac) ] / (2a), where a = 3, b = 80, and c = -8000:

x = [ -80 ± √(80² - 4(3)(-8000)] / (2×3)
  x = [ -80 ± √(6400 + 96000) ] / 6
  x = [ -80 ± √102400 ] / 6
  x = [ -80 ± 320 ] / 6

We discard the negative solution since x represents the number of units sold, leaving:

x = (-80 + 320) / 6 = 240 / 6 = 40

Thus, the revenue-maximizing number of units is 40.

Now, find the maximum revenue by substituting x = 40 back into R(x):

R(40) = 8000(40) - 40(40)² - (40)³
  R(40) = 320,000 - 40(1600) - 64,000
  R(40) = 320,000 - 64,000 - 64,000
  R(40) = 192,000

Therefore, to maximize revenue, 40 units must be sold, resulting in a maximum revenue of 192,000.
To maximize revenue, sell 40 units, which yields a maximum revenue of 192,000.

If total revenue for a firm is given by

where is the number of units sold, find the number of units that must be sold to
maximize revenue. Find the maximum revenue.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Calculus Questions

If total revenue for a firm is given by where is the number of units sold, find the number of units that must be sold to maximize revenue. Find the maximum revenue.