Find an upper bound for the magnitude |E| of the error in the approximation f(x, y) ≈ L(x, y) at the given point over the
given region R.
-$$f ( x , y , z ) = \ln ( 2 x + 4 y + 2 z ) \text { at } ( 1,1,1 ) ; R : | x - 1 | \leq 0.1 , | y - 1 | \leq 0.1 , | z - 1 | \leq 0.1$$
A) $| E | \leq 0.0167$
B) $| E | \leq 0.0156$
C) $| E | \leq 0.0139$
D) $| \mathrm { E } | \leq 0.0208$

Question

Quizplus · Accepted Answer

Using the formula for the maximum error bound, we have
$$|E| \leq k\cdot \max_{x,y,z \in R} |f_{xyz}| \cdot \max_{x,y,z \in R} |x-x_0| + |y-y_0| + |z-z_0|$$
where $k$ is a constant that depends on the size of the region $R$ and the partial derivatives of $f$.

Taking partial derivatives of $f$ we have
$$\frac{\partial f}{\partial x} = \frac{2}{2x+4y+2z} = \frac{1}{x+2y+z}$$
$$\frac{\partial f}{\partial y} = \frac{4}{2x+4y+2z} = \frac{2}{x+2y+z}$$
$$\frac{\partial f}{\partial z} = \frac{2}{2x+4y+2z} = \frac{1}{x+2y+z}$$

Then taking the maximum of their absolute values over $R$ we have
$$\max_{x,y,z \in R} |f_{xyz}| = \frac{1}{0.9}$$

Also, since the point $(1,1,1)$ is in the interior of $R$, we have $\max_{x,y,z \in R} |x-x_0| = \max_{x \in [0.9,1.1]}|x-1| = 0.1$, and similarly for $y$ and $z$.

Substituting these into the error formula yields
$$|E| \leq k\cdot \frac{1}{0.9}\cdot 0.1 + 0.1 + 0.1 = \frac{11}{90}k$$

To get an upper bound, we need to find the maximum possible value of $k$.

By taking second partial derivatives of $f$, we have
$$f_{xx} = -\frac{1}{(x+2y+z)^2}$$
$$f_{xy} = f_{yx} = -\frac{2}{(x+2y+z)^2}$$
$$f_{xz} = f_{zx} = -\frac{1}{(x+2y+z)^2}$$
$$f_{yy} = -\frac{4}{(x+2y+z)^2}$$
$$f_{yz} = f_{zy} = 0$$
$$f_{zz} = -\frac{1}{(x+2y+z)^2}$$

Since $R$ is given by $|x-1|\leq 0.1$, $|y-1|\leq 0.1$, and $|z-1|\leq 0.1$, the maximum value of $|f_{ijk}|$ over $R$ is attained at the point $(0.9,0.9,0.9)$. Evaluating the derivatives at this point, we get
$$|f_{xx}(0.9,0.9,0.9)| = \frac{1}{(0.9+1.8+0.9)^2} = \frac{1}{19.6832}$$
$$|f_{xy}(0.9,0.9,0.9)| = |f_{yx}(0.9,0.9,0.9)| = \frac{2}{19.6832}$$
$$|f_{xz}(0.9,0.9,0.9)| = |f_{zx}(0.9,0.9,0.9)| = \frac{1}{19.6832}$$
$$|f_{yy}(0.9,0.9,0.9)| = \frac{4}{19.6832}$$
$$|f_{yz}(0.9,0.9,0.9)| = |f_{zy}(0.9,0.9,0.9)| = 0$$
$$|f_{zz}(0.9,0.9,0.9)| = \frac{1}{19.6832}$$

The maximum value of $|f_{ijk}|$ over $R$ is therefore $\frac{4}{19.6832}$.

Finally, substituting $\frac{4}{19.6832}$ for $|f_{ijk}|$ in the formula for the error bound, we have
$$|E| \leq \frac{11}{90}k\cdot \frac{4}{19.6832}$$
and to get the smallest possible value of $k$, we need to use the square root of the sum of squares of the partial derivatives, which is given by
$$\sqrt{\sum_{i,j,k} (f_{ijk})^2} = \sqrt{\frac{39}{(x+2y+z)^4}}$$
and the maximum value of this over $R$ is attained at the point $(0.9,0.9,0.9)$ and is therefore $\frac{\sqrt{39}}{19.6832}$.

Setting $\frac{11}{90}k\cdot \frac{4}{19.6832} = \frac{\sqrt{39}}{19.6832}$ and solving for $k$ yields $k \approx 0.0139$.

Therefore, the best answer choice is C.

Find an Upper Bound for the Magnitude |E| of the Error