(a) Finding the area of a square chip with side length $x$:

The area of a square is given by the formula: $A(x) = x^2 \text{ square millimeters (sq mm)}$

This gives the area of a square chip in terms of its side length $x$ in millimeters.

(b) Finding $A'(11)$ and its meaning:

To find the rate of change of the area with respect to the side length $x$, we need to take the derivative of $A(x) = x^2$:

$A'(x) = \frac{d}{dx}(x^2) = 2x$

Now, to find $A'(11)$, substitute $x = 11$ into the derivative:

$A'(11) = 2(11) = 22 \text{ square millimeters per millimeter (sq mm/mm)}$

Interpretation:

$A'(11) = 22$ means that when the side length of the wafer is 11 mm, the area of the square increases at a rate of 22 square millimeters for every additional millimeter increase in side length.

(c) Finding the rate of change of area when the perimeter is 52 mm:

The perimeter $P$ of a square is given by: $P = 4x$

To find the side length $x$ when the perimeter is 52 mm: $52 = 4x \implies x = \frac{52}{4} = 13 \text{ mm}$

Now, we compute $A'(x)$ at $x = 13$: $A'(x) = 2x$

Thus, $A'(13) = 2(13) = 26 \text{ square millimeters per millimeter (sq mm/mm)}$

So, the rate of change of the area when the perimeter is 52 mm is $26 \, \text{sq mm/mm}$.

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Do you want more details on any step? Here are 5 questions to deepen your understanding:

1. How is the derivative formula $A'(x) = 2x$ derived?
2. What is the geometric meaning of $A'(x)$ for different values of $x$?
3. How would the rate of change change for smaller or larger values of side length?
4. What if the side length were kept close to another value, say 15 mm?
5. Can this approach be applied to other shapes, such as circles?

Tip: The derivative of an area function like $A(x)$ always represents how fast the area changes as the side length changes.