The function is given as,
[tex]f(x,y)=7x^2+2y^2-7x+16y-13[/tex]Obtain partial derivative with respect to 'x' as,
[tex]\begin{gathered} \frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(7x^2+2y^2-7x+16y-13) \\ \frac{\partial f}{\partial x}=(14x+0-7+0-0) \\ \frac{\partial f}{\partial x}=14x-7 \end{gathered}[/tex]Obtain partial derivative with respect to 'y' as,
[tex]\begin{gathered} \frac{\partial f}{\partial y}=\frac{\partial}{\partial y}(7x^2+2y^2-7x+16y-13) \\ \frac{\partial f}{\partial y}=(0+4y-0+16-0) \\ \frac{\partial f}{\partial y}=4y+16 \end{gathered}[/tex]The critical points can be obtained by equating the partial derivatives to zero,
[tex]\begin{gathered} \frac{\partial f}{\partial x}=0\Rightarrow14x-7=0\Rightarrow x=0.5 \\ \frac{\partial f}{\partial y}=0\Rightarrow4y+16=0\Rightarrow y=-4 \end{gathered}[/tex]So the critical point of the function is (0.5,-4).
Obtain the second-order derivatives as,
[tex]\begin{gathered} r=\frac{\partial^2f}{\partial x^2}=\frac{\partial}{\partial x}(14x-7)=14-0=14 \\ s=\frac{\partial^2f}{\partial x\partial y^{}}=\frac{\partial}{\partial x}(4y+16)=0+0=0 \\ t=\frac{\partial^2f}{\partial y^2}=\frac{\partial}{\partial y}(4y+16)=4+0=4 \end{gathered}[/tex]Consider the following criteria,
1. Relative Maximum:
[tex]rt-s^2>0\text{ and }r>0\Rightarrow\text{ Relative minima }[/tex]2. Relative Minimum:
[tex]rt-s^2>0\text{ and }r<0\Rightarrow\text{ Relative maxima }[/tex]3. Saddle Point:
[tex]rt-s^2<0\Rightarrow\text{ Saddle Point }[/tex]4. No Conclusion can be drawn:
[tex]rt-s^2=0\Rightarrow\text{ Inconclusive }[/tex]Check for the given values of second derivatives,
[tex]\begin{gathered} r(0.5,-4)=14 \\ s(0.5,-4)=0 \\ t(0.5,-4)=4 \end{gathered}[/tex]It follows that,
[tex]\begin{gathered} rt-s^2=14(4)-(0)^2=14(4)>0 \\ r=14>0 \end{gathered}[/tex]Thus, the function has a relative minimum at the critical point (0.5,-4).
The value of this maximum can be obtained by substituting the critical point coordinates in the function,
[tex]\begin{gathered} f_{\max }=f(0.5,-4) \\ f_{\max }=7(0.5)^2+2(-4)^2-7(0.5)+16(-4)-13 \\ f_{\max }=1.75+32-3.5-64-13 \\ f_{\max }=-46.75 \end{gathered}[/tex]Thus, the maximum value of the function is - 46.75 which occurs at the point (0.5,-4).
