Chapter 4.6, Problem 11E

To determine

To calculate: Thelargest possible volume of the box if $1200$ ${\text{cm}}^2$ of material is available to make a box with a square base and an open top.

Answer to Problem 11E

The largest possible volume of the box is $4000$ ${\text{cm}}^3$ .

Explanation of Solution

Given information:

Material available to make a box with a square base and an open top is $1200$ ${\text{cm}}^2$

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. $f^{″}(a)>0$ 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

Material available to make a box with a square base and an open top is $1200$ ${\text{cm}}^2$

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

The surface area of open box is $A=b^2+4bh$

  $b^2+4bh=1200$

Solve the equation for $h$ , to get

  $\begin{array}{l}4bh=1200-b^2\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}h=\frac{1200}{4b}-\frac{b^2}{4b}\\ h=\frac{300}{b}-\frac{b}{4}(1)\end{array}$

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 $V=b^{2}h$

Use equation $(1)$ , to get

  $V(b)=b^2\left(\frac{300}{b}-\frac{b}{4}\right)$

Simplified the volume equation,

  $\begin{array}{l}V(b)=b^2\left(\frac{300}{b}-\frac{b}{4}\right)\\ V(b)=b^2\times \frac{300}{b}-b^2\times \frac{b}{4}\\ V(b)=300b-\frac{b^3}{4}\end{array}$

Differentiate on both sides,

  $V^{\prime }(b)=300-\frac{3b^2}{4}\mathrm{......}\left(2\right)$

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

  $300-\frac{3b^2}{4}=0$

Subtract $300$ on both sides,

  $-\frac{3b^2}{4}=-300$

Multiplied by $-4$ on both sides,

  $3b^2=1200$

Divide by $3$ on both sides,

  $b^2=400$

Take the square root on both sides,

  $b=\pm \sqrt{400}$

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

Therefore,

  $\begin{array}{l}b=\sqrt{400}\\ b=20\end{array}$

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

  $\begin{array}{l}{V}^{″}(b)=-\frac{3b^2}{4}\\ V^{″}(b)=-\frac{3\times 2\times b}{4}\\ V^{″}(b)=-\frac{3b}{2}\end{array}$

Substitute $b=20$ and simplified

  $\begin{array}{l}{V}^{″}(20)=-\frac{3\times 20}{2}\\ V^{″}(20)=-3\times 10\\ V^{″}(20)=-30\end{array}$

Therefore,

  $V^{″}(20)=-30\le 0$

For $b=20$ volume of box is maximize

Substitute $b=20$ in equation $(1)$ and simplified

  $\begin{array}{l}h=\frac{300}{20}-\frac{20}{4}\\ h=15-5\\ h=10\end{array}$

Now, put $h=10$ , $b=20$ in volume equation $V=b^{2}h$ , to get

  $\begin{array}{l}V={20}^2\times 10\\ V=400\times 10\\ V=4000\end{array}$

Thus the largest possible volume of the box is $4000$ ${\text{cm}}^3$