The weekly revenue for a product is given by ​R(x)equals118.8 x minus 0.0105 x squared​, and the weekly cost is ​C(x)equals11 comma 000 plus 59.4 x minus 0.021 x squared plus 0.00001 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?
Question content area bottom
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.)
Part 2
​(b) The maximum possible profit is ​$
  
enter your response here.

Answer 1

To find the maximum profit, we need to first express the profit function, which is given by:

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

Step 1: Define the Revenue and Cost Functions

The revenue function is:

\[ R(x) = 118.8x - 0.0105x^2 \]

The cost function is:

\[ C(x) = 11000 + 59.4x - 0.021x^2 + 0.00001x^3 \]

Step 2: Define the Profit Function

The profit function \(P(x)\) can be defined as:

\[ P(x) = (118.8x - 0.0105x^2) - (11000 + 59.4x - 0.021x^2 + 0.00001x^3) \]

Simplifying the expression:

\[ P(x) = 118.8x - 0.0105x^2 - 11000 - 59.4x + 0.021x^2 - 0.00001x^3 \] \[ P(x) = (118.8 - 59.4)x + (-0.0105 + 0.021)x^2 - 0.00001x^3 - 11000 \] \[ P(x) = 59.4x + 0.0105x^2 - 0.00001x^3 - 11000 \]

Step 3: Find the Maximum Profit

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

\[ P'(x) = 59.4 + 0.021x - 0.00003x^2 \]

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

\[ 0.00003x^2 - 0.021x - 59.4 = 0 \]

Step 4: Use the Quadratic Formula

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

Here, \(a = 0.00003\), \(b = -0.021\), \(c = -59.4\).

Calculating the discriminant:

\[ D = (-0.021)^2 - 4(0.00003)(-59.4) \] \[ D = 0.000441 + 0.007128 \] \[ D = 0.007569 \]

Now calculate \(x\):

\[ x = \frac{-(-0.021) \pm \sqrt{0.007569}}{2 \times 0.00003} \] \[ x = \frac{0.021 \pm 0.08693}{0.00006} \]

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

1. \(x_1 = \frac{0.021 + 0.08693}{0.00006} \approx 1801.67\)
2. \(x_2 = \frac{0.021 - 0.08693}{0.00006} \approx -1.097\) (not feasible)

Step 5: Rounding and Finding the Maximum Profit

Rounding \(x_1\) to the nearest whole number gives:

\[ x \approx 1802 \]

Step 6: Calculate Maximum Profit \(P(1802)\)

Substituting \(x = 1802\) into the profit function:

\[ P(1802) = 59.4(1802) + 0.0105(1802^2) - 0.00001(1802^3) - 11000 \]

Calculating each term:

1. \(59.4 \times 1802 = 107133.6\)
2. \(0.0105 \times 1802^2 = 0.0105 \times 3247204 \approx 34064.15\)
3. \(-0.00001 \times 1802^3 \approx -0.00001 \times 5836216808 \approx -58362.17\)
4. Subtracting 11000.

Now, calculate total profit:

\[ P(1802) = 107133.6 + 34064.15 - 58362.17 - 11000 \]

Calculating gives:

\[ P(1802) \approx 107133.6 + 34064.15 - 58362.17 - 11000 \approx 57535.58 \]

Final Results

Part 1: (a) The number of units that will give the maximum profit is 1802.

Part 2: (b) The maximum possible profit is approximately $57535.58.