We are given the formula for the surface area $S$ as:

$S = 0.121 w^{0.31} h^{0.903}$

where:

• $w$ is the weight in pounds,
• $h$ is the height in inches,
• $S$ is the surface area in square feet.

The goal is to estimate the maximum error in the surface area $S$ when the errors in the measurements of $w$ and $h$ are at most 2%.

Step 1: Relative Error in Surface Area

The relative error in $S$, denoted by $\frac{\Delta S}{S}$, can be approximated using partial derivatives of $S$ with respect to $w$ and $h$.

The total differential of $S$ is:

$dS = \frac{\partial S}{\partial w} dw + \frac{\partial S}{\partial h} dh$

Dividing both sides by $S$, we get the relative error:

$\frac{dS}{S} = \frac{\partial S}{\partial w} \cdot \frac{dw}{S} + \frac{\partial S}{\partial h} \cdot \frac{dh}{S}$

This formula gives us the relative change in $S$ in terms of the relative changes in $w$ and $h$.

Step 2: Calculate Partial Derivatives

We first calculate the partial derivatives:

$\frac{\partial S}{\partial w} = 0.121 \cdot 0.31 \cdot w^{0.31 - 1} h^{0.903} = 0.03751 w^{-0.69} h^{0.903}$

$\frac{\partial S}{\partial h} = 0.121 \cdot 0.903 \cdot w^{0.31} h^{0.903 - 1} = 0.10926 w^{0.31} h^{-0.097}$

Step 3: Relative Changes in $w$ and $h$

Given that the relative error in $w$ and $h$ is at most 2%, we have:

$\frac{dw}{w} = 0.02 \quad \text{and} \quad \frac{dh}{h} = 0.02$

Step 4: Maximum Relative Error in $S$

The relative error in $S$ is approximated by:

$\frac{dS}{S} \approx 0.31 \cdot \frac{dw}{w} + 0.903 \cdot \frac{dh}{h}$

Substitute $\frac{dw}{w} = 0.02$ and $\frac{dh}{h} = 0.02$:

$\frac{dS}{S} \approx 0.31 \cdot 0.02 + 0.903 \cdot 0.02$

$\frac{dS}{S} \approx 0.0062 + 0.01806 = 0.02426$

So the relative error in $S$ is approximately 0.02426, or 2.426%.

Step 5: Maximum Error in $S$

Thus, the maximum error in the calculated surface area is approximately 2.426%.

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Would you like more details or have any questions about the process?

Here are 5 related questions to expand your understanding:

1. How would the error change if the relative error in $w$ was larger than 2%?
2. Can we use a different method to estimate the error for nonlinear models?
3. How does increasing the height $h$ impact the surface area in terms of sensitivity to error?
4. What happens if the measurement errors in $w$ and $h$ are not independent?
5. How would the result change if we also had a percentage error in the constant 0.121?

Tip: In error estimation, relative error gives a good approximation for small errors, especially when dealing with multiplication and power functions.