Questions: When no more than 110 units are produced, the cost of producing y units is given by C(x). C(x) = 0.2x^3 - 25x^2 + 1520x + 27,231 How many units should be produced in order to have the lowest possible average cost? units should be produced. (Round to one decimal place as needed.)

When no more than 110 units are produced, the cost of producing y units is given by C(x).
C(x) = 0.2x^3 - 25x^2 + 1520x + 27,231

How many units should be produced in order to have the lowest possible average cost?
units should be produced.
(Round to one decimal place as needed.)

Transcript text: When no more than 110 units are produced, the cost of producing $y$ units is given by $\mathrm{C}(\mathrm{x})$. \[ C(x)=0.2 x^{3}-25 x^{2}+1520 x+27,231 \] How many units should be produced in order to have the lowest possible average cost? $\square$ units should be produced. (Round to one decimal place as needed.)

Solution

Solution Steps

To find the number of units that should be produced to achieve the lowest possible average cost, we need to minimize the average cost function. The average cost function is given by the total cost function $ C(x) $ divided by the number of units $ x $. We will first derive the average cost function, then find its derivative, and solve for the critical points by setting the derivative to zero. Finally, we will evaluate these points to determine which one gives the minimum average cost.

Step 1: Understand the Problem

We are given a cost function $ C(x) = 0.2x^3 - 25x^2 + 1520x + 27,231 $ and need to find the number of units $ x $ that should be produced to minimize the average cost. The average cost function is given by:

\[ AC(x) = \frac{C(x)}{x} = \frac{0.2x^3 - 25x^2 + 1520x + 27,231}{x} \]

Step 2: Simplify the Average Cost Function

Simplify the average cost function by dividing each term by $ x $:

\[ AC(x) = 0.2x^2 - 25x + 1520 + \frac{27,231}{x} \]

Step 3: Find the Derivative of the Average Cost Function

To find the minimum average cost, we need to find the derivative of $ AC(x) $ and set it to zero:

\[ AC'(x) = \frac{d}{dx}\left(0.2x^2 - 25x + 1520 + \frac{27,231}{x}\right) \]

Using the power rule and the derivative of $ \frac{1}{x} $, we get:

\[ AC'(x) = 0.4x - 25 - \frac{27,231}{x^2} \]

Step 4: Set the Derivative to Zero and Solve for $ x $

Set the derivative equal to zero to find the critical points:

\[ 0.4x - 25 - \frac{27,231}{x^2} = 0 \]

Multiply through by $ x^2 $ to clear the fraction:

\[ 0.4x^3 - 25x^2 - 27,231 = 0 \]

This is a cubic equation that can be solved using numerical methods or graphing techniques. For simplicity, we will use a numerical approach to find the approximate value of $ x $.

Step 5: Solve the Cubic Equation Numerically

Using a numerical solver or graphing calculator, we find that the approximate solution to the equation $ 0.4x^3 - 25x^2 - 27,231 = 0 $ is:

\[ x \approx 110.5 \]

Step 6: Round to One Decimal Place

The problem asks us to round to one decimal place, so we have:

\[ x = 110.5 \]