Chapter 14, Problem 1CE

To determine

To prove: The critical point correspondence to a minimum of the provided function

$f\left(m,b\right)=3b^2+11,460,644m^2+11,724bm-51.86b-101,460m+225.045$.

Explanation of Solution

Given:

The critical points are: $m=0.01$ and $b=-10.85$. The provided function is:

$f\left(m,b\right)=3b^2+11,460,644m^2+11,724bm-51.86b-101,460m+225.045$.

Formula Used:

For minimum, the condition needs to be true.

$\frac{{\partial }^{2}f}{\partial m^2}\times \frac{{\partial }^{2}f}{\partial b^2}-{\left[\frac{\delta \left(\frac{\delta f}{\delta m}\right)}{\delta b}\right]}^2>0$

Proof:

Provided function is:

$f\left(m,b\right)=3b^2+11,460,644m^2+11,724bm-51.86b-101,460m+225.045$

Calculate $\frac{{\partial }^{2}f}{\partial m^2}$ for the function:

$\frac{\delta f}{\delta m}=22,921,288m+11,724b-101,460$

And $\frac{\begin{array}{l}\\ {\partial }^{2}f\end{array}}{\partial m^2}=22,921,288$

Calculate $\frac{{\partial }^{2}f}{\partial b^2}$ for the function:

$\frac{\delta f}{\delta b}=6b+11,724m-101,460$

And $\frac{\begin{array}{l}\\ {\partial }^{2}f\end{array}}{\partial b^2}=6$.

Calculate $\frac{\delta \left(\frac{\delta f}{\delta m}\right)}{\delta b}$ for the function:

$\begin{array}{l}\frac{\delta f}{\delta m}=22,921,288m+11,724b-101,460\\ \\ \text{and}\\ \frac{\begin{array}{l}\\ \partial \left(\frac{\delta f}{\delta m}\right)\end{array}}{\partial b}=11,724\end{array}$

Calculate $\frac{{\partial }^{2}f}{\partial m^2}\times \frac{{\partial }^{2}f}{\partial b^2}-{\left[\frac{\delta \left(\frac{\delta f}{\delta m}\right)}{\delta b}\right]}^2$ for the function:

$\begin{array}{ccc}\hfill \frac{{\partial }^{2}f}{\partial m^2}\times \frac{{\partial }^{2}f}{\partial b^2}-{\left[\frac{\delta \left(\frac{\delta f}{\delta m}\right)}{\delta b}\right]}^2& =& 22,921,288\times 6-{\left(11,724\right)}^2\hfill \\ \hfill & =& 137,527,728-137,452,176\hfill \\ \hfill & =& 75,552\hfill \end{array}$

Hence, $\frac{{\partial }^{2}f}{\partial m^2}\times \frac{{\partial }^{2}f}{\partial b^2}-{\left[\frac{\delta \left(\frac{\delta f}{\delta m}\right)}{\delta b}\right]}^2>0$

Therefore, provided critical point correspondence to a minimum of the provided function.