Chapter 5.4, Problem 10E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 10E

The minimum length of the fence the dimensions of pea patch are $\left(12m,18m\right)$ and the length of the fence is $72m$ .

Explanation of Solution

Given information: Given The area of the pea patch is $216m^2$ .

Calculation:

Let the sides length of the subdivide fence is $x$

Dimension of the pea patch is $xby\frac{216}{x}m$

Let the total fence required is $F$

  $\begin{array}{l}F=\left(3x\right)+2\left(\frac{216}{x}\right)\\ =3x+432x^{-1}\end{array}$

For any function $f\left(x\right)$ the condition of minimum and maximum is obtained by find $f^{\prime }\left(x\right)=0$ ,

Critical points and end points

  $\begin{array}{l}\mathrm{<mo>F</mo>}=3x+432x^{-1}\\ \frac{dF}{dx}=3x-432x^{-2}\end{array}$

Critical points occur at

  $\begin{array}{l}\text{ <msup><mo> F</mo> <mo>′</mo></msup>}=0\\ 3x-432x^{-2}=0\\ x^2=\frac{432}{3}\\ =144\\ x=\pm 12\end{array}$

  $\text{Since }0<x<12\text{ consider }x=12$

solve it graphically, it has critical point occur at $x=12,$

Here $\frac{dF}{dx}>0\text{ for 0< }x<12\text{and}\frac{dF}{dx}>0\text{ for }x>12$ this corresponds to minimum length of the fence.

For

  $\begin{array}{l}x=12\\ \frac{216}{12}=18\end{array}$

Required length of the fence is

  $\begin{array}{l}F=3x+2\left(\frac{216}{x}\right)\\ =36+36\\ =72\end{array}$

Graphical support:

  

Thus, the minimum length of the fence the dimensions of pea patch are $\left(12m,18m\right)$ and the length of the fence is $72m$ .