Question

Cost of Manufacturing The following data represent the cost C (in thousands of dollars) of manufacturing Chevy Cobalts and the number x of Cobalts produced.

(a) Draw a scatter diagram of the data using x as the independent variable and C as the dependent variable. Comment on the type of relation that may exist between the two variables C and x.

(b) Use a graphing utility to find the cubic function of best-fit C = C(x).

(c) Graph the cubic function of best fit on the scatter diagram.

(d) Use the function found in part (b) to predict the cost of manufacturing 11 Cobalts.

(e) Interpret the y-intercept.

Short answer

Part (a)

Part (b) The obtained model is $C\left(x\right)=0.2156x^3-2.3473x^2+14.328x+10.224$.

Part (c)

Part (d) The obtained result is $\begin{array}{rcl}& & \$170.7723\mathrm{thousends}\end{array}$.

Part (e) The obtained cubic model is best-fit for the given data.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given table.

Part (a) Step 2. Scatter plot. 

Draw the scatter plot by using the given table.

Part (b) Step 1. Find the function.

Use the graphing utility to find the best-fit cubic function.

The result is$C\left(x\right)=0.2156x^3-2.3473x^2+14.328x+10.224$and the graph is shown below,

Part (c) Step 1. Draw the graph on the scatter plot.

The cubic function of best fit on the scatter diagram is given below,

Part (d) Step 1. Find the value at x=11.

Substitute 11 for x in the obtained function.

$\begin{array}{rcl}C\left(11\right)& =& 0.2156{\left(11\right)}^3-2.3473{\left(11\right)}^2+14.328\left(11\right)+10.224\\ & =& \$170.7723\mathrm{thousends}\end{array}$

Part (e) Step 5. Find the y-intercept.

Substitute 0 for x in the obtained function. Use the calculator to find the value.

$\begin{array}{rcl}C\left(0\right)& =& 0.2156{\left(0\right)}^3-2.3473{\left(0\right)}^2+14.328\left(0\right)+10.224\\ & =& 0+10.224\\ & =& 10.224\end{array}$

Thus, when the value of no cobalt is 0 then the cost is approx 10 thousand dollars. The obtained cubic model is best fit for the given data.