Chapter 4.6, Problem 6E

To determine

To calculate: Thedimensions of a rectangle with Area $1000$${\text{m}}^2$ whose perimeter is as small as possible.

Answer to Problem 6E

The rectangle with minimum perimeter is square whose side length is $31.6$$\text{m}$ .

Explanation of Solution

Given information:

The Area of the rectangle whose perimeter is as small as possible is $1000$${\text{m}}^2$ .

Formula used:

Let $l$ be the length and $b$ be the breadth of rectangle. Then

1. The perimeter of rectangle is $P=2(l+b)$ .
2. Area of rectangle is $A=l\times b$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$ , if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$ , if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

The Area of rectangle $=$

  $1000$${\text{m}}^2$

Let the length be $l$ and breadth be $b$ .

Recall that,

The Ares of rectangle is $A=l\times b$ .

  $l\times b=1000$

Divide by $l$ on both sides

  $b=\frac{1000}{l}\mathrm{.............}(1)$

Recall that,

The Perimeter of rectangle is $P=2(l+b)$

Simplify,

  $P=2l+2b$

Use equation $(1)$

  $P=2l+2\times \frac{1000}{l}$

Simplify the equation,

  $P=2l+\frac{2000}{l}$

Recall that,

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$ , if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate on both sides,

  $P^{\prime }(l)=2-2000\left(\frac{1}{l^2}\right)\mathrm{......}\left(2\right)$

Solve for $P^{\prime }(l)=0$ ,

  $2-2000\left(\frac{1}{l^2}\right)=0$

Subtract $2$ on both sides,

  $-\frac{2000}{l^2}=-2$

Multiplied by $-l^2$ on both sides,

  $2000=2l^2$

Divided by $2$ on both sides,

  $1000=l^2$

Take the square root on both sides, to get

  $l=\pm \sqrt{1000}$

The equation has two real solution but length can’t be negative,

Therefore,

  $\begin{array}{l}l=\sqrt{1000}\\ l=31.6(approx)\end{array}$

Thus the rectangle with minimum perimeter is square whose side length is $31.6$$\text{m}$