The weekly revenue for a product is given by ​R(x)equals356.4 x minus 0.0585 x squared​, and the weekly cost is ​C(x)equals10 comma 000 plus 178.2 x minus 0.117 x squared plus 0.00003 x cubed​, where x is the number of units produced and sold.

​(a) How many units will give the maximum​ profit?
​(b) What is the maximum possible​ profit?
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Part 1
​(a) The number of units that will give the maximum profit is
  
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1 answer

To find the number of units that will give the maximum profit, we first need to define the profit function \( P(x) \) using the revenue function \( R(x) \) and the cost function \( C(x) \):

\[ P(x) = R(x) - C(x) \]

Given that:

\[ R(x) = 356.4x - 0.0585x^2 \] \[ C(x) = 10,000 + 178.2x - 0.117x^2 + 0.00003x^3 \]

We can substitute these into the profit function:

\[ P(x) = (356.4x - 0.0585x^2) - (10,000 + 178.2x - 0.117x^2 + 0.00003x^3) \]

Simplifying the profit function:

\[ P(x) = 356.4x - 0.0585x^2 - 10,000 - 178.2x + 0.117x^2 - 0.00003x^3 \]

Combining like terms gives:

\[ P(x) = (356.4 - 178.2)x + (-0.0585 + 0.117)x^2 - 0.00003x^3 - 10,000 \] \[ P(x) = 178.2x + 0.0585x^2 - 0.00003x^3 - 10,000 \]

To find the maximum profit, we take the derivative of \( P(x) \) and set it to zero:

\[ P'(x) = 178.2 + 0.117x - 0.00009x^2 \]

Setting \( P'(x) = 0 \) to find the critical points:

\[ 0 = 178.2 + 0.117x - 0.00009x^2 \]

Rearranging gives the quadratic equation in standard form:

\[ 0.00009x^2 - 0.117x - 178.2 = 0 \]

We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:

  • \( a = 0.00009 \)
  • \( b = -0.117 \)
  • \( c = -178.2 \)

Calculating \( b^2 - 4ac \):

\[ b^2 = (-0.117)^2 = 0.013689 \] \[ 4ac = 4 \cdot 0.00009 \cdot (-178.2) = -0.064176 \] Thus,

\[ b^2 - 4ac = 0.013689 + 0.064176 = 0.077865 \]

Now we can compute \( x \):

\[ x = \frac{-(-0.117) \pm \sqrt{0.077865}}{2 \cdot 0.00009} \] \[ x = \frac{0.117 \pm 0.2795}{0.00018} \]

Calculating the two possible values for \( x \):

  1. \( x = \frac{0.117 + 0.2795}{0.00018} = \frac{0.3965}{0.00018} \approx 2203 \)
  2. \( x = \frac{0.117 - 0.2795}{0.00018} = \frac{-0.1625}{0.00018} \) (this value will be negative and not relevant)

Thus, the number of units that will give the maximum profit is:

\[ \boxed{2203} \]

Part 2: To find the maximum possible profit, we can substitute \( x = 2203 \) back into the profit function \( P(x) \):

Calculating \( R(2203) \):

\[ R(2203) = 356.4(2203) - 0.0585(2203^2) \]

Calculating \( C(2203) \):

\[ C(2203) = 10,000 + 178.2(2203) - 0.117(2203^2) + 0.00003(2203^3) \]

Finally, find \( P(2203) = R(2203) - C(2203) \) to get the maximum profit.

Performing the above calculations will yield the maximum profit value.