91影视

It is given that function is

$f(x,y)=1.2-0.2x^2-0.3y^2+0.1xy-0.25x$

Points are $(0.5,-0.5)$

The gradient of $f(x,y)$is given by

$鈭�f(x,y)=f_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j}+f_z(x,y)\mathrm{k}$

$鈭�f(x,y)=f_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j}$

role="math" localid="1653829678282" $=\frac{鈭�}{鈭�x}\left(1.2-0.2x^2-0.3y^2+0.1xy-0.25x\right)\mathrm{i}+\frac{鈭�}{鈭�y}\left(1.2-0.2x^2-0.3y^2\right.+0.1xy-0.25x)\mathrm{j}$

$=(-0.4x+0.1y-0.25)\mathrm{i}+(-0.6y+0.1x)\mathrm{j}$

At $(0.5,-0.5)$

$鈭�f(0.5,-0.5)=鉄�-0.5,0.35鉄�$is increasing function

Thus function decreases in direction $-鈭�f(0.5,-0.5)=-鉄�-0.5,0.35鉄�$

Hence, required points are

$鉄�0.5,-0.35鉄�$

It is given by$D_{\mathbf{u}}f\left(x_0,y_0\right)=鈭�f\left(x_0,y_0\right)路\mathbf{u}$

The direction derivative at given direction is zero since it neither ascend not descend

$D_{\mathbf{u}}f\left(x_0,y_0\right)=鈭�f\left(x_0,y_0\right)路\mathbf{u}$

$=0$

Hence, gradient of given function at $(0.5,-0.5)$is

$鈭�f(0.5,-0.5)=鉄�-0.5,0.35鉄�$

Assuming be unit vector

$鈭�f\left(x_0,y_0\right)路\mathbf{u}=0$

$-u_1+0.7u_2=0$

$u_1{}^2+u_2{}^2=1$

Using $u_1=0.7u_2$in above equation

${\left(0.7u_2\right)}^2+u_2^2=1$

$u_2=卤\frac{1}{\sqrt{1.49}}$

Using $u_2=卤\frac{1}{\sqrt{1.49}}\text{in}{u}_1=0.7u_2$

$u_1=卤\frac{0.7}{\sqrt{1.49}}$

The required direction is