Chapter 4.4, Problem 18E

(a.)

To determine

A formula y $V\left(x\right)$ for the volume of the box.

Answer to Problem 18E

It has been determined that the formula for the volume of the box is e $V\left(x\right)=\left(\frac{15}{2}-x\right)\left(10-2x\right)x$ .

Explanation of Solution

Given:

A piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$

side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid, as shown in the following figure:

  

Concept used:

The volume of a cuboid is the product of length, width and height.

Calculation:

It is given that a piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$ side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid.

So, the shape of the box so formed is that of a cuboid.

It can be seen from the figure that the length of the box is the length of the base, which is $NK$ .

Now, from the figure, $AN+NK+KJ+JI=15in.$

Again, from the figure, $AN=KJ=x$ and since the dimensions of the lid must equal the dimensions of the base, it follows that $JI=NK$ .

Put these values in $AN+NK+KJ+JI=15in.$ to get,

  $x+NK+x+NK=15in.$

Simplifying,

  $2\left(NK+x\right)=15$

On further simplification,

  $NK=\frac{15}{2}-x$

So, the length of the box is $\frac{15}{2}-x$ .

Similarly, it can be seen from the figure that the width of the box is the width of the base, which is $CN$ .

Now, from the figure, $DC+CN+NM=10in.$

Again, from the figure, $DC=NM=x$ .

Put these values in $DC+CN+NM=10in.$ to get,

  $x+CN+x=10in.$

Simplifying,

  $CN+2x=10$

On further simplification,

  $CN=10-2x$

So, the width of the box is $10-2x$ .

Finally, it can be seen from the figure that the height of the box is given as $NM$ .

Now, from the figure, $NM=x$ .

So, the height of the box is $x$ .

As determined, the length, width and height of the box are $\frac{15}{2}-x$ , $10-2x$ and $x$ respectively.

Then, the volume of the cuboidal box is given as,

  $V\left(x\right)=\left(\frac{15}{2}-x\right)\left(10-2x\right)x$

This is the required formula for volume of the box.

Conclusion:

It has been determined that the formula for the volume of the box is $V\left(x\right)=\left(\frac{15}{2}-x\right)\left(10-2x\right)x$ .

(b.)

To determine

The domain of $V\left(x\right)$ in the given problem and the graph of $V\left(x\right)$ over its domain.

Answer to Problem 18E

It has been determined that the domain of $V\left(x\right)$ in the given problem is $0<x<5$ and the graph of $V\left(x\right)$ over its domain is given as follows:

  

Explanation of Solution

Given:

A piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$

side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid, as shown in the following figure:

  

Concept used:

The domain of a function $V\left(x\right)$ is the set of values of $x$ for which the function is defined.

Calculation:

It is given that a piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$ side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid.

So, the shape of the box so formed is that of a cuboid.

As determined previously, the length, width and height of the box are $\frac{15}{2}-x$ , $10-2x$ and $x$ respectively.

Since the dimensions of the box must be positive, it follows that $\frac{15}{2}-x>0$ , $10-2x>0$ and $x>0$ .

Simplifying, $x<\frac{15}{2}$ , $2x<10$ and $x>0$ .

On further simplification, $x<7.5$ , $x<5$ and $x>0$ .

Combining, $0<x<5$ .

This is the domain of $V\left(x\right)$ in the given problem.

As determined previously, the volume of the cuboidal box is given as,

  $V\left(x\right)=\left(\frac{15}{2}-x\right)\left(10-2x\right)x$

The graph of $V\left(x\right)$ over its domain; $0<x<5$ is given as:

  

Conclusion:

It has been determined that the domain of $V\left(x\right)$ in the given problem is $0<x<5$ and the graph of $V\left(x\right)$ over its domain is given as follows:

  

(c.)

To determine

The maximum volume and the value of $x$ that gives it, using graphical method.

Answer to Problem 18E

It has been determined that the maximum volume is approximately $66.019$ cubic inches and it is given by $x\approx 1.92$ .

Explanation of Solution

Given:

A piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$

side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid, as shown in the following figure:

  

Concept used:

The highest point on a graph gives the maximum value of the function.

Calculation:

It is given that a piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$ side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid.

So, the shape of the box so formed is that of a cuboid.

As determined previously, the volume of the cuboidal box is given as,

  $V\left(x\right)=\left(\frac{15}{2}-x\right)\left(10-2x\right)x$

The graph of $V\left(x\right)$ over its domain; $0<x<5$ is given as:

  

It can be seen that the highest point on the graph of $V\left(x\right)$ is the point $\left(1.92,66.019\right)$ .

This implies that the maximum volume is approximately $66.019$ cubic inches and it is given by $x\approx 1.92$ .

Conclusion:

It has been determined that the maximum volume is approximately $66.019$ cubic inches and it is given by $x\approx 1.92$ .

(d.)

To determine

To Confirm: The result in part (c) analytically.

Answer to Problem 18E

The result obtained in part (c) has been confirmed analytically.

Explanation of Solution

Given:

A piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$

side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid, as shown in the following figure:

  

Concept used:

A function attains its maximum value at a point where its first derivative is zero and its second derivative is negative.

Calculation:

It is given that a piece of cardboard measures $10in.$ by $15in.$ Two equal squares are removed from the corners of a $10in.$ side and two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with a lid.

So, the shape of the box so formed is that of a cuboid.

As determined previously, the volume of the cuboidal box is given as,

  $V\left(x\right)=\left(\frac{15}{2}-x\right)\left(10-2x\right)x$

Differentiating,

  $V^{\prime }\left(x\right)=\left(-1\right)\left(10-2x\right)x+\left(\frac{15}{2}-x\right)\left(-2\right)x+\left(\frac{15}{2}-x\right)\left(10-2x\right)\left(1\right)$

Simplifying,

  $\begin{array}{c}{V}^{\prime }\left(x\right)=x\left(2x-10\right)+\left(\frac{15}{2}-x\right)\left(-2x+10-2x\right)\\ =2x^2-10x+\left(\frac{15}{2}-x\right)\left(10-4x\right)\\ =2x^2-10x+\frac{15}{2}\left(10-4x\right)-x\left(10-4x\right)\\ =2x^2-10x+75-30x-10x+4x^2\end{array}$

So, $V^{\prime }\left(x\right)=6x^2-50x+75$ .

Differentiating again,

  $V^{″}\left(x\right)=12x-50$

Equating the first derivative to zero,

  $6x^2-50x+75=0$

Applying the quadratic formula,

  $x=\frac{-\left(-50\right)\pm \sqrt{{\left(-50\right)}^2-4\left(6\right)\left(75\right)}}{2\left(6\right)}$

Solving,

  $\begin{array}{c}x=\frac{50\pm \sqrt{2500-1800}}{12}\\ =\frac{50\pm \sqrt{700}}{12}\\ =\frac{50\pm 10\sqrt{7}}{12}\\ =\frac{25\pm 5\sqrt{7}}{6}\end{array}$

So, $x=\frac{25+5\sqrt{7}}{6}$ or $x=\frac{25-5\sqrt{7}}{6}$ .

Simplifying, $x\approx 6.37$ or $x\approx 1.96$ .

Now, it has been determined previously that the domain is $0<x<5$ .

So, the only feasible solution is $x\approx 1.96$ .

Put $x\approx 1.96$ in $V^{″}\left(x\right)=12x-50$ to get,

  $V^{″}\left(1.96\right)\approx -26.48<0$

This shows that the value of $V\left(x\right)$ is maximum when $x\approx 1.96$ .

This confirms the result obtained in part (c).

Conclusion:

The result obtained in part (c) has been confirmed analytically.