Given $$ f(x y)-16-x^{2}-y^{2} $$, use the total differential to approximate $$ \Delta z $$ at $$ (3,2) $$ towards $$ (2,2,1.02) $$. Round your answer to two decimal places.

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Let
$$
z=f(x,y)=xy-16-x^2-y^2.
$$
We wish to use the total differential at the point $$(3,2)$$ to approximate the change $$\Delta z$$ when moving from $$(x,y)=(3,2)$$ to $$(x,y)=(2,2)$$. (The given $$1.02$$ in the point $$(2,2,1.02)$$ is the new $$z$$–value.)

A linear (total‐differential) approximation is given by
$$
\Delta z\approx dz=f_x(x,y)\,\Delta x+f_y(x,y)\,\Delta y,
$$
where
$$
\Delta x=2-3=-1\quad\text{and}\quad\Delta y=2-2=0.
$$

Step 1. Compute the partial derivatives.

Differentiate $$f(x,y)=xy-16-x^2-y^2$$:

- The partial derivative with respect to $$x$$ is
  $$
  f_x(x,y)=\frac{\partial}{\partial x}\Bigl(xy-16-x^2-y^2\Bigr)=y-2x.
  $$
  
- The partial derivative with respect to $$y$$ is
  $$
  f_y(x,y)=\frac{\partial}{\partial y}\Bigl(xy-16-x^2-y^2\Bigr)=x-2y.
  $$

Step 2. Evaluate these derivatives at $$(3,2)$$.

- At $$(3,2)$$
  $$
  f_x(3,2)=2-2\cdot3=2-6=-4,
  $$
  $$
  f_y(3,2)=3-2\cdot2=3-4=-1.
  $$

Step 3. Apply the total differential formula.

We have $$\Delta x=-1$$ and $$\Delta y=0$$; therefore,
$$
dz=f_x(3,2)(\Delta x)+f_y(3,2)(\Delta y)=(-4)(-1)+(-1)(0)=4+0=4.
$$

Thus, the approximate change in $$z$$ when moving from $$(3,2)$$ to $$(2,2)$$ is
$$
\Delta z\approx4.00.
$$
The approximate change in $$ z $$ is 4.00.

Given , use the total differential to approximate at towards . Round
your answer to two decimal places.

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Given , use the total differential to approximate at towards . Round your answer to two decimal places.