Chapter 4.1, Problem 39PPSE

Solution

To state: The number of chips that minimum the cost if the equation for the cost in dollars of producing computer chips is $C=0.000015x^2-0.03x+35$ , where x is the number of chips produced. Also find the cost for that number o chips.

The number of chips that minimize the cost is 1000 and minimum cost is $20.

Given information:

The given equation is $C=0.000015x^2-0.03x+35$ .

Explanation:

Consider the equation:

  $C=0.000015x^2-0.03x+35$

Rewrite the equation as shown below:

  $\begin{array}{c}C=0.000015x^2-0.03x+35\\ =0.000015\left(x^2-2000x\right)+35\\ =0.000015\left(x^2-2\cdot 1000\cdot x+{1000}^2-{1000}^2\right)+35\\ =0.000015\left({\left(x-1000\right)}^2-1000000\right)+35\\ =0.000015{\left(x-1000\right)}^2-15+35\\ =0.000015{\left(x-1000\right)}^2+20\end{array}$

The equation of the parabola is $C=0.000015{\left(x-1000\right)}^2+20$ .

Now compare the equation $C=0.000015{\left(x-1000\right)}^2+20$ with $y=a{\left(x-h\right)}^2+k$ .

The value of a is 0.000015 and that of k is 20.

Now the number of chips that minimize the cost is:

  $C=0.000015{\left(x-1000\right)}^2+20$

Substitute the value of C as 20,

  $\begin{array}{l}20=0.000015{\left(x-1000\right)}^2+20\\ 0=0.000015{\left(x-1000\right)}^2\\ 0={\left(x-1000\right)}^2\\ x=1000\end{array}$

Therefore, the number of chips that minimize the cost is 1000 and their minimum cost is $20.