Chapter 4.6, Problem 12E

To determine

To calculate: Thedimension of a box that minimize the amount of material used of the box with a square base and open top must have a volume of $32000{\text{cm}}^3$

Answer to Problem 12E

The side length of the square box and height of the box are respectively $\text{40}\text{cm}$ and $\text{20}\text{cm}$ .

Explanation of Solution

Given information:

A box with a square base and open top must have a volume of $32000{\text{cm}}^3$

Formula used:

Let $b$ be the length of a side of a square base and $h$ be the height of open boxThen

1. The surface area of open box is $A=b^2+4bh$
2. Volume of open box $V=b^{2}h$

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.
4. 1. If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   2. If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

The volume material available to make a box with a square base and an open top is $32000{\text{cm}}^3$

Recall that, let $b$ be the length of a side of a square base and $h$ be the height of open box Then

Volume of open box $V=b^{2}h$

  $b^{2}h=32000$

Divide by $b^2$ on both sides

  $h=\frac{32000}{b^2}(1)$

Recall that, the surface area of open box is $A=b^2+4bh$

Use equation $(1)$

  $A(b)=b^2+4b\times \frac{32000}{b^2}$

Simplify the equation,

  $\begin{array}{l}A(b)=b^2+4b\times \frac{32000}{b^2}\\ A(b)=b^2+4\times \frac{32000}{b}\\ A(b)=b^2+\frac{128000}{b}\end{array}$

Recall that,

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.
4. 1. $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   2. If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $A^{\prime }(b)=2b-\frac{128000}{b^2}\mathrm{......}\left(2\right)$

Solve for $A^{\prime }(b)=0$ ,

  $2b-\frac{128000}{b^2}=0$

Add $\frac{128000}{b^2}$ on both sides,

  $2b=\frac{128000}{b^2}$

Multiplied by $b^2$ on both sides,

  $2b^3=128000$

Divided by $2$ on both sides

  $b^3=64000$

Take the cube root on both sides

  $\begin{array}{l}b=\sqrt[3]{64000}\\ b=\sqrt[3]{40\times 40\times 40}\\ b=40\end{array}$

Differentiate equation $\left(2\right)$ with respect to $b$

  $A^{″}(b)=2+\frac{256000}{b^3}$

Substitute $b=40$ and simplified

  $\begin{array}{l}{A}^{″}(40)=2+\frac{256000}{{40}^3}\\ A^{″}(40)=2+\frac{256000}{40\times 40\times 40}\\ A^{″}(40)=2+4\\ A^{″}(40)=6\end{array}$

Therefore,

  $A^{″}(b)=6\ge 0$

For $b=40$ volume of box is minimize

Substitute $b=40$ in equation $(1)$

  $\begin{array}{l}h=\frac{32000}{{40}^2}\\ h=\frac{32000}{40\times 40}\\ h=20\end{array}$

Thus the side length of the square box and height of the box are respectively $\text{40}\text{cm}$ and $\text{20}\text{cm}$