Question

attempt 1: the answer is incorrect. 9 attempts remaining. suppose ( c(x) = 0.03x^2 + 2x + 3800 ) is the total cost of a company to produce ( x ) units of a certain product. find the production level ( x ) that minimizes the average cost ( a(x) = \frac{c(x)}{x} ). (round the production level to two decimal places.)

Explanation:

Step1: Define Average Cost Function

The average cost function \( A(x) \) is given by \( A(x)=\frac{C(x)}{x} \). Substitute \( C(x) = 0.03x^{2}+2x + 3800 \) into the formula:

$$ A(x)=\frac{0.03x^{2}+2x + 3800}{x}=0.03x + 2+\frac{3800}{x} $$

Step2: Find Derivative of \( A(x) \)

To find the minimum, we take the derivative of \( A(x) \) with respect to \( x \).
The derivative of \( 0.03x \) is \( 0.03 \), the derivative of \( 2 \) is \( 0 \), and the derivative of \( \frac{3800}{x}=3800x^{-1} \) is \( - 3800x^{-2}=-\frac{3800}{x^{2}} \). So:

$$ A^{\prime}(x)=0.03-\frac{3800}{x^{2}} $$

Step3: Set Derivative to Zero and Solve for \( x \)

To find critical points, set \( A^{\prime}(x) = 0 \):

$$ 0.03-\frac{3800}{x^{2}}=0 $$

Add \( \frac{3800}{x^{2}} \) to both sides:

$$ 0.03=\frac{3800}{x^{2}} $$

Multiply both sides by \( x^{2} \):

$$ 0.03x^{2}=3800 $$

Divide both sides by \( 0.03 \):

$$ x^{2}=\frac{3800}{0.03}=\frac{380000}{3}\approx126666.67 $$

Take the square root of both sides (we consider positive \( x \) since \( x \) represents production level):

$$ x=\sqrt{\frac{380000}{3}}\approx\sqrt{126666.67}\approx355.90 $$

Step4: Verify Minimum (Second Derivative Test)

Find the second derivative of \( A(x) \). The derivative of \( A^{\prime}(x)=0.03 - 3800x^{-2} \) is \( A^{\prime\prime}(x)=\frac{7600}{x^{3}} \). For \( x>0 \), \( A^{\prime\prime}(x)>0 \), so the function is concave up at this critical point, meaning it is a minimum.

Answer:

\( 355.90 \)