Minimizing the Average Cost of Producing a Commercial Product Suppose that [latex]C(x)=0.02 x^2+2 x+4000[/latex] is the total cost (in dollars) for a company to produce x units of a certain product. Find the production level x that minimizes the average cost.

The average cost function is given by [latex]\bar{C}(x)=\frac{0.02 x^2+2 x+4000}{x}=0.02 x+2+4000 x^{-1} .[/latex] To minimize [latex]\bar{C}(x)[/latex], we start by finding critical numbers in the domain x > 0. We have [latex]\bar{C}^{\prime}(x)=0.02-4000 x^{-2}=0 \quad \text { if }[/latex] [latex]4000 x^{-2}=0.02 \text { or }[/latex] [latex]\frac{4000}{0.02}=x^2[/latex]. Then [latex]x^2=200,000 \text { or } x=\pm \sqrt{200,000} \approx \pm 447[/latex]. Since x > 0, the only relevant critical number is at approximately x = 447. Further, [latex]\bar{C}^{\prime}(x)<0 \text { if } x<447[/latex] and [latex]\bar{C}^{\prime}(x)>0 \text { if } x>447[/latex], so this critical number is the location of the absolute minimum on the domain x > 0. A graph of the average cost function (see Figure 3.97) shows the minimum.

Question 3.9.2: Minimizing the Average Cost of Producing a Commercial Produc...

Calculus: Early Transcendental Functions

Minimizing the Average Cost of Producing a Commercial Product

Suppose that

C(x)=0.02 x^2+2 x+4000

is the total cost (in dollars) for a company to produce x units of a certain product. Find the production level x that minimizes the average cost.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

The average cost function is given by

\bar{C}(x)=\frac{0.02 x^2+2 x+4000}{x}=0.02 x+2+4000 x^{-1} .

To minimize $\bar{C}(x)$ , we start by finding critical numbers in the domain x > 0. We have

\bar{C}^{\prime}(x)=0.02-4000 x^{-2}=0 \quad \text { if }

$4000 x^{-2}=0.02 \text { or }$

$\frac{4000}{0.02}=x^2$ .

Then $x^2=200,000 \text { or } x=\pm \sqrt{200,000} \approx \pm 447$ . Since x > 0, the only relevant critical number is at approximately x = 447. Further, $\bar{C}^{\prime}(x)<0 \text { if } x<447$ and $\bar{C}^{\prime}(x)>0 \text { if } x>447$ , so this critical number is the location of the absolute minimum on the domain x > 0. A graph of the average cost function (see Figure 3.97) shows the minimum.