he 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.

Question

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 express the profit \( P(x) \) in terms of revenue \( R(x) \) and cost \( C(x) \):

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

Substituting the given revenue and cost functions:

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

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

Now, we can express the profit function:

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

Simplifying this, we group like terms:

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

This gives:

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

Calculating the coefficients:

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

Next, we want to find the maximum profit by determining the critical points of \( P(x) \). We do this by taking the derivative of \( P(x) \) and setting it to zero:

\[ \frac{dP}{dx} = 178.2 + 0.117x - 0.00009x^2 \]

Setting the derivative equal to zero:

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

This is a quadratic equation in the standard form:

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

To make it easier to solve, let's multiply everything by -1:

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

We can now use the quadratic formula where \( a = 0.00009 \), \( b = -0.117 \), and \( c = -178.2 \):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

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

\[ b^2 = (-0.117)^2 = 0.013689 \] \[ 4ac = 4 \cdot 0.00009 \cdot -178.2 = -0.06467 , (\text{approximately}) \] \[ b^2 - 4ac = 0.013689 - (-0.06467) \approx 0.013689 + 0.06467 = 0.078359 \]

Now substituting the values back into the formula:

\[ x = \frac{0.117 \pm \sqrt{0.078359}}{0.00018} \]

Calculating the square root and solving for \( x \):

\[ \sqrt{0.078359} \approx 0.280 \]

This gives two possible solutions:

\[ x_1 = \frac{0.117 + 0.280}{0.00018} \approx \frac{0.397}{0.00018} \approx 2205.56 \] \[ x_2 = \frac{0.117 - 0.280}{0.00018} \approx \frac{-0.163}{0.00018} \approx -905.56 , (\text{Not valid since x cannot be negative}) \]

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

\[ x \approx 2206 , (\text{rounded to the nearest whole number}) \]

The final answer is:

\[ \boxed{2206} \]