Chapter 4.4, Problem 48E

a.

To determine

To find the equation of the square of length.

Answer to Problem 48E

  $L^2=\frac{2x^3}{(2x-8.5)}$

Explanation of Solution

Given:

The dimensions of the rectangle sheet is $\frac{81}{2}\times 11\text{ in}$ .

Calculation:

Considering the figure, by Pythagoras theorem

The square of the hypotenuse is equal to the sum of square of adjacent side and opposite side.

Let QR=y and PB=8.5-x

  

  $\begin{array}{l}QB=\sqrt{x^2-{(8.5-x)}^2}\\ QB=\sqrt{x^2-({8.5}^2+x^2-2(x)(8.5))}\\ QB=\sqrt{x^2-{8.5}^2-x^2+2(x)(8.5))}\\ QB=\sqrt{-{8.5}^2+2(x)(8.5))}\end{array}$

Here triangles QRS and PQB are similar. Therefore

  $\begin{array}{l}\frac{QR}{RS}=\frac{PQ}{QB}\\ \frac{y}{8.5}=\frac{x}{\sqrt{8.5(2x-8.5)}}\\ \frac{y^2}{{8.5}^2}=\frac{x^2}{8.5(2x-8.5)}\\ \frac{y^2}{8.5}=\frac{x^2}{(2x-8.5)}\\ y^2=\frac{8.5x^2}{(2x-8.5)}\\ L^2=x^2+y^2\\ L^2=x^2+\frac{8.5x^2}{(2x-8.5)}\\ L^2=\frac{x^2(2x-8.5)+8.5x^2}{2x-8.5}\\ L^2=\frac{2x^3-8.5x^2+8.5x^2}{2x-8.5}\\ L^2=\frac{2x^3}{2x-8.5}\end{array}$

b.

To determine

To find the minimum of $L^2$ .

Answer to Problem 48E

The $L^2$ is minimum at $x=6.37$

Explanation of Solution

Given:

The dimensions of the rectangle sheet is $\frac{81}{2}\times 11\text{ in}$ .

Calculation:

Let

  $\begin{array}{l}f(x)=L^2\\ f(x)=\frac{2x^3}{2x-8.5}\end{array}$

For any function $f(x)$ the condition of minimum and maximum is obtained by finding $f\text{'}(x)=0$ , critical points and end points

  $\begin{array}{l}f(x)=L^2\\ f(x)=\frac{2x^3}{2x-8.5}\\ \frac{df(x)}{dx}=\frac{d}{dx}\left(\frac{2x^3}{2x-8.5}\right)\\ \frac{df(x)}{dx}=\frac{(2x-8.5)(6x^2)-(2x^3)(2)}{{(2x-8.5)}^2}\\ \frac{df(x)}{dx}=\frac{(12x^3-51x^2)-(4x^3)}{{(2x-8.5)}^2}\\ \frac{df(x)}{dx}=\frac{(8x^3-51x^2)}{{(2x-8.5)}^2}\end{array}$

Critical point occurs at

  $\begin{array}{l}f\text{'}(x)=\frac{(8x^3-51x^2)}{{(2x-8.5)}^2}\\ 0=\frac{(8x^3-51x^2)}{{(2x-8.5)}^2}\\ (8x^3-51x^2)=0\\ 8x^3=51x^2\\ x=\frac{51}{8}\\ x=6.37\end{array}$

C.

To determine

To find the minimum value of $L^2$ .

Answer to Problem 48E

The minimum value of $L^2$ is $11.04$ in.

Explanation of Solution

Given:

The strength S of a rectangle wooden beam is proportional to its width times the square of its depth.

  $S=kw{d}^3$

Where S,w,d,k are the strength, width, depth and a constant respectively.

Calculation:

  $L^2=\frac{2x^3}{2x-8.5}$

At $x=6.37$ , the minimum value is found.

Substitute $x=6.37$ in $L^2$

  $\begin{array}{l}{L}^2=\frac{2{(6.37)}^3}{2(6.37)-8.5}\\ L^2=\frac{2(258.474853)}{12.74-8.5}\\ L^2=\frac{516.949706}{4.24}\\ L^2=121.9221004\\ L=11.04\end{array}$