The weekly revenue for a product is given by ​R(x)equals396 x minus 0.0405 x squared​, and the weekly cost is ​C(x)equals12 comma 000 plus 198 x minus 0.081 x squared plus 0.00003 x cubed​, where x is the number of units produced and sold.

Question

​(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
  
enter your response here.
​(Round to the nearest whole number as​ needed.)

Answer 1

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

1. Profit Function: \[ P(x) = R(x) - C(x) \]

   Given: \[ R(x) = 396x - 0.0405x^2 \] \[ C(x) = 12,000 + 198x - 0.081x^2 + 0.00003x^3 \]

   Substitute \( R(x) \) and \( C(x) \) into the profit function: \[ P(x) = (396x - 0.0405x^2) - (12,000 + 198x - 0.081x^2 + 0.00003x^3) \]

   Simplifying this: \[ P(x) = 396x - 0.0405x^2 - 12,000 - 198x + 0.081x^2 - 0.00003x^3 \] \[ P(x) = (396 - 198)x + (-0.0405 + 0.081)x^2 - 0.00003x^3 - 12,000 \] \[ P(x) = 198x + 0.0405x^2 - 0.00003x^3 - 12,000 \]

   Rearranging gives: \[ P(x) = -0.00003x^3 + 0.0405x^2 + 198x - 12,000 \]

2. Finding Maximum Profit: To find the maximum profit, we take the derivative of \( P(x) \) and set it to zero: \[ P'(x) = -0.00009x^2 + 0.081x + 198 \]

   Setting \( P'(x) = 0 \): \[ -0.00009x^2 + 0.081x + 198 = 0 \]

   To simplify the calculations, we can multiply through by \( -1000000 \) (to eliminate the decimals): \[ 90x^2 - 81000x - 198000000 = 0 \]

   Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

   • \( a = 90 \)
   • \( b = -81000 \)
   • \( c = -198000000 \)

   The discriminant \( D \) is: \[ D = (-81000)^2 - 4(90)(-198000000) \] \[ D = 6561000000 + 7128000000 \] \[ D = 13629000000 \]

   Now calculating \( x \): \[ x = \frac{81000 \pm \sqrt{13629000000}}{2 \times 90} \] \[ x = \frac{81000 \pm 116766.77}{180} \]

   Calculating the positive root: \[ x \approx \frac{198766.77}{180} \approx 1104.26 \]

   Rounding to the nearest whole number, we get: \[ x = 1104 \]

(a) The number of units that will give the maximum profit is 1104.

Now for (b) The maximum possible profit:

We substitute \( x = 1104 \) back into \( P(x) \) to find the maximum profit: \[ P(1104) = -0.00003(1104)^3 + 0.0405(1104)^2 + 198(1104) - 12,000 \] Calculating each term:

• \( (1104)^3 \approx 133,100,544 \)
• \( (1104)^2 \approx 1,219,216 \)

Now substituting: \[ P(1104) = -0.00003(133100544) + 0.0405(1219216) + 198(1104) - 12,000 \] Calculating:

1. \( -0.00003 \cdot 133100544 \approx -3993.02 \)
2. \( 0.0405 \cdot 1219216 \approx 49300.88 \)
3. \( 198 \cdot 1104 = 218592 \)

Adding these together: \[ P(1104) \approx -3993.02 + 49300.88 + 218592 - 12000 \] \[ P(1104) \approx 206899.86 \]

Thus, rounding gives:

(b) The maximum possible profit is approximately 206900.