a company that produces cell phones has a cost function of $c = 27x^2 - 580x + 18977$, where $c$ is the cost in dollars and $x$ is the number of cell phones produced (in thousands). how many units of cell phone (in thousands) minimizes the cost function?

$x =$ thousand phones produced will minimize $c$

Question

a company that produces cell phones has a cost function of \$c = 27x^2 - 580x + 18977\$, where \$c\$ is the cost in dollars and \$x\$ is the number of cell phones produced (in thousands). how many units of cell phone (in thousands) minimizes the cost function?

\$x =\$ thousand phones produced will minimize \$c\$

$x =$   thousand phones produced will minimize $c$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Critical Points", "Absolute Extrema on Closed Intervals" ], "new_concepts": [ "Cost Minimization" ], "current_concepts": [ "Critical Points", "Absolute Extrema on Closed Intervals", "Cost Minimization" ] } ## Find the derivative of the cost function $$ \begin{aligned} C(x) &= 27x^2 - 580x + 18977 \ C'(x) &= 54x - 580 \end{aligned} $$ ## Find the critical point by setting the derivative to zero $$ \begin{aligned} 54x - 580 &= 0 \ 54x &= 580 \ x &= \frac{580}{54} = \frac{290}{27} \approx 10.74 \end{aligned} $$ ## Verify minimization using the second derivative test $$ \begin{aligned} C''(x) &= 54 > 0 \end{aligned} $$ A company that produces cell phones has a cost function of $C = 27x^2 - 580x + 18977$, where $C$ is the cost in dollars and $x$ is the number of cell phones produced (in thousands). How many units of cell phone (in thousands) minimizes the cost function? $x =$ $10.74$ thousand phones produced will minimize $C$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Optimization Problems" ] }

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QUESTION IMAGE

Question

Explanation:

Find the derivative of the cost function

Find the critical point by setting the derivative to zero

Verify minimization using the second derivative test

Answer: