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hoa [83]
3 years ago
5

A certain electronics manufacturer found that the average cost C to produce x DVD/Blu- ray players can be found using the equati

on C=0.03x2−7x+800. What is the minimum average cost per machine and how many DVD/Blu-ray players should be built in order to acheive that minimum?
Mathematics
1 answer:
lilavasa [31]3 years ago
5 0

Answer:

<em>The minimum average cost is $2.80 when 163 players are built</em>

Step-by-step explanation:

<u>Average Cost Function</u>

<em>We'll assume the given function as the total cost to produce x players, and NOT the average cost since that is a different definition, as shown below.</em>

The cost C to produce x DVD/Blu- ray players is given by the equation

C=0.03x^2-7x+800

The Average Cost function  is defined as

\displaystyle\bar  C=\frac{C}{x}

\displaystyle\bar  C=\frac{0.03x^2-7x+800}{x}

\displaystyle\bar  C=0.03x-7+\frac{800}{x}

To find the extreme value of the average cost, we must take the first derivative of the function

\displaystyle\bar  C'=0.03-\frac{800}{x^2}

Equating to 0

\displaystyle 0.03-\frac{800}{x^2}=0

Solving for x

\displaystyle x=\sqrt{\frac{800}{0.03}}

x=163

The second derivative is

\displaystyle\bar  C''=\frac{1600}{x^3}

For x=163 the second derivative is positive, thus x=163 is a minimum value. Let's compute the minimum average cost

\displaystyle\bar  C(163)=0.03\cdot 163-7+\frac{800}{163}

\displaystyle\bar  C(163)=2.80

The minimum average cost is $2.80

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