Chapter 4.4, Problem 28E

a.

To determine

Find the values at which the absolute maximum of $f$ occurs.

Answer to Problem 28E

The absolute maxima of function $f$ occurs at $x=1$ .

Explanation of Solution

Given:

The function $f(x)=x{e}^{-x}$ .

Concept Used:

Calculation:

Function: $f(x)=x{e}^{-x}$

Differentiate the $f(x)$ ,

  $\begin{array}{c}\frac{d}{dx}\left(f(x)\right)=\frac{d}{dx}\left(x{e}^{-x}\right)\\ f\text{'}(x)=e^{-x}-x{e}^{-x}\\ f\text{'}(x)=e^{-x}\left(1-x\right).\end{array}$

For absolute maxima put $f\text{'}(x)=0$ ,

  $\begin{array}{c}f\text{'}(x)=0\\ e^{-x}\left(1-x\right)=0\\ \left(1-x\right)=0\\ x=1.\end{array}$

So, absolute maxima of function $f$ occurs at $x=1$ .

Conclusion:

The absolute maxima of function $f$ occurs at $x=1$ .

b.

To determine

Complete the table.

Answer to Problem 28E

$a$	$b$	$A$
0.1	3.7150	$0.3270$
0.2	2.8650	0.4356
0.3	2.3646	0.4589
0.4	2.0188	0.4341
0.5	1.7564	0.3809
0.6	1.5474	0.3118
0.7	1.3755	0.2346
0.8	1.2308	0.1549
0.9	1.1071	0.0757
1	1	0

Explanation of Solution

Given:

The function $f(x)=x{e}^{-x}$ .

Graph of function $f(x)$:

  

Calculation:

According to graph,

  $f(a)=f(b)$

  $a=0.1$

Then,

  $\begin{array}{c}f(x)=x{e}^{-x}\\ \text{put }x=a=0.1\\ f(a)=0.1e^{-0.1}\\ =0.090484\\ \text{But, }f(a)=f(b)\\ f(a)=0.090484\\ f(b)=0.090484\\ b=3.7150\end{array}$

So, the graph is

  

Area of rectangle $A$:

  $\begin{array}{c}A=\text{ length × width}\\ A=3.6150\times 0.0905\\ A=0.3270\end{array}$

Area of rectangle is $A=0.3270$ square unit.

Similarly, find the other values. The table is

$a$	$b$	$A$
0.1	3.7150	$0.3270$
0.2	2.8650	0.4356
0.3	2.3646	0.4589
0.4	2.0188	0.4341
0.5	1.7564	0.3809
0.6	1.5474	0.3118
0.7	1.3755	0.2346
0.8	1.2308	0.1549
0.9	1.1071	0.0757
1	1	0

c.

To determine

To Draw: a scatter plot of the data $(a,A)$ .

Answer to Problem 28E

  

Explanation of Solution

Given:

$a$	$b$	$A$
0.1	3.7150	$0.3270$
0.2	2.8650	0.4356
0.3	2.3646	0.4589
0.4	2.0188	0.4341
0.5	1.7564	0.3809
0.6	1.5474	0.3118
0.7	1.3755	0.2346
0.8	1.2308	0.1549
0.9	1.1071	0.0757
1	1	0

Calculation:

  

Here, $x-axis$ denotes values of $a$ and $y-axis$ denotes $A$ .

d.

To determine

To Find the quadratic, cubic and quartic regression equations for the data in part (b).

Answer to Problem 28E

The absolute maxima of function $f$ occurs at $x=1$ .

Explanation of Solution

Given:

The table is

$a$	$A$
0.1	$0.3270$
0.2	0.4356
0.3	0.4589
0.4	0.4341
0.5	0.3809
0.6	0.3118
0.7	0.2346
0.8	0.1549
0.9	0.0757
1	0

Concept used:

Quadratic regression equation:

  $A=p+qa+r{a}^2$

Here,

  $\begin{array}{l}r=\frac{\left[{\displaystyle \sum a^{2}A\ast }{\displaystyle \sum aa}\right]-\left[{\displaystyle \sum aA\ast }{\displaystyle \sum a{a}^2}\right]}{\left[{\displaystyle \sum aa\ast }{\displaystyle \sum A^2{A}^2}\right]-{\left[{\displaystyle \sum a{a}^2}\right]}^2}\\ q=\frac{\left[{\displaystyle \sum aA\ast }{\displaystyle \sum a^2{a}^2}\right]-\left[{\displaystyle \sum a^{2}A\ast }{\displaystyle \sum a{a}^2}\right]}{\left[{\displaystyle \sum aa\ast }{\displaystyle \sum A^2{A}^2}\right]-{\left[{\displaystyle \sum a{a}^2}\right]}^2}\\ p=\left[{\displaystyle \sum \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right]-\left\{b\ast \left[{\displaystyle \sum \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right]\right\}-\left\{a\ast \left[{\displaystyle \sum \raisebox{1ex}{$a^2$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right]\right\}\end{array}$

Calculation:

Quadratic regression equation:-

  $A=p+qa+r{a}^2$

Here,

  $\begin{array}{l}r=\frac{\left[{\displaystyle \sum a^{2}A\ast }{\displaystyle \sum aa}\right]-\left[{\displaystyle \sum aA\ast }{\displaystyle \sum a{a}^2}\right]}{\left[{\displaystyle \sum aa\ast }{\displaystyle \sum A^2{A}^2}\right]-{\left[{\displaystyle \sum a{a}^2}\right]}^2}\\ q=\frac{\left[{\displaystyle \sum aA\ast }{\displaystyle \sum a^2{a}^2}\right]-\left[{\displaystyle \sum a^{2}A\ast }{\displaystyle \sum a{a}^2}\right]}{\left[{\displaystyle \sum aa\ast }{\displaystyle \sum A^2{A}^2}\right]-{\left[{\displaystyle \sum a{a}^2}\right]}^2}\\ p=\left[{\displaystyle \sum \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right]-\left\{b\ast \left[{\displaystyle \sum \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right]\right\}-\left\{a\ast \left[{\displaystyle \sum \raisebox{1ex}{$a^2$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\right]\right\}\end{array}$

On solving,

  $A=-0.911a^2+0.539a+0.336$

Similarly, find the cubic and quartic regression equations.

Cubic regression equations

  $A=0.1865+1.8632a-3.7827a^2+1.7402a^3$ .

Quartic regression equations

  $A=0.1204+2.7109a-6.8847a^2+5.9789a^3-1.9267a^4$ .

Conclusion:

Quadratic regression equation

  $A=-0.911a^2+0.539a+0.336$ .

Cubic regression equations

  $A=0.1865+1.8632a-3.7827a^2+1.7402a^3$ .

Quartic regression equations

  $A=0.1204+2.7109a-6.8847a^2+5.9789a^3-1.9267a^4$ .

d.

To determine

Find maximum possible value of the area using the regression equations.

Answer to Problem 28E

From quadratic regression equation,

Maximum possible area of the rectangle is $0.4157$ square units.

From cubic regression equation,

Maximum possible area of the rectangle is $0.4525$ square units.

From quartic regression equation,

Maximum possible area of the rectangle is $0.4599$ square units.

Explanation of Solution

Given:

Quadratic regression equation

  $A=-0.911a^2+0.539a+0.336$ .

Cubic regression equations

  $A=0.1865+1.8632a-3.7827a^2+1.7402a^3$ .

Quartic regression equations

  $A=0.1204+2.7109a-6.8847a^2+5.9789a^3-1.9267a^4$ .

Calculation:

Draw the graphs of regression equations,

  

Now, according to graph,

From quadratic regression equation,

Maximum possible area of the rectangle is $0.4157$ square units.

From cubic regression equation,

Maximum possible area of the rectangle is $0.4525$ square units.

From quartic regression equation,

Maximum possible area of the rectangle is $0.4599$ square units.

Conclusion:

For quadratic regression equation: $\text{Max}\text{. Area}=0.4157$

For cubic regression equation: $\text{Max}\text{. Area}=0.4525$

For quartic regression equation: $\text{Max}\text{. Area}=0.4599$