Chapter 3.8, Problem 11E

(a)

To determine

To find: The value of $A^{\prime }\left(15\right)$.

Answer to Problem 11E

The value of $A^{\prime }\left(15\right)$ is $\underset{\_}{30{\text{mm}}^{\text{2}}}$.

Explanation of Solution

Given:

The company makes square wafers of silicon of area of the form $A\left(x\right)=x^2$.

Calculation:

Differentiate the area of the square wafer respect to x.

$A\left(x\right)=x^2$

$A\text{'}\left(x\right)=2x$

Substitute 15 for $x$ in the above equation and find $A^{\prime }\left(15\right)$ as,

$\begin{array}{c}A\text{'}\left(15\right)=2\left(15\right)\\ =30{\text{mm}}^{\text{2}}\end{array}$

Therefore, the rate at which the area will increase with respect to change in its side length, if the side length is 15 mm will be $\underset{\_}{30{\text{mm}}^{\text{2}}}$.

(b)

To determine

To Show: The rate of change of the area of a square with respect to the change in its side length is equal to half its perimeter.

Explanation of Solution

Draw the square with side length, x as shown in Figure (1).

The perimeter of the square is $P\left(x\right)=4x$.

From the figure 1, the change in length $\Delta x$ is very small.

The first derivative of $A\left(x\right)$ is,

$\begin{array}{c}A\text{'}\left(x\right)=2x\\ =\frac{1}{2}\left(4x\right)\\ =\frac{1}{2}P\left(x\right)\end{array}$

Thus, the change in area of the square is approximately half of its perimeter which is half of the 4 sides times.

The expression for the shaded area (change in area) as below.

$\Delta A=2x\left(\Delta x\right)+{\left(\Delta x\right)}^2$

The change in length $\Delta x$ is small, and change in area $\Delta A$ is approximately equal to $2x\left(\Delta x\right)$.

$\Delta A=2x\left(\Delta x\right)$

The term $\frac{\Delta A}{\Delta x}$ is approximately equal to $2x$.

$\frac{\Delta A}{\Delta x}=2x$

$\frac{\Delta A}{\Delta x}=\frac{1}{2}P\left(x\right)$

Thus the rate of change of the area of a square with respect to the change in its side length is equal to half its perimeter is $\frac{\Delta A}{\Delta x}=\frac{1}{2}P\left(x\right)$.