< Previous Section | Home | Next Section > |

Figure 3.13.1 shows a 38 cm × 32 cm rectangular sheet of cardboard. You can fold the sheet into a box by cutting a square with side lengths

We want to define a function that gives the box’s volume in relation to the side length of the square cutouts.

The volume of a rectangular box is

Let

Let

Let

Let

Let

Now we can use Figure 3.13.1 to guide us in defining the functions

$V(x) = h(x) \cdot A(x)$ | The box’s volume in relation to a
value of x is its height in relation to x times the area of its
base in relation to the value of x. |

$h(x) = x$ | The box’s height in relation to a value
of x is equal to the length of the cutouts’ sides, which is x. |

$A(x) = l(x) \cdot w(x)$ | The area of the box’s base in relation to a value of x is the
base’s width in relation to the value of x times its length in
relation to the value of x. |

$w(x) = 32 - 2x$ | The box’s width in relation to a value of x is 32 cm minus
twice the cutouts’ width. |

$l(x) = 38 - 2x$ | The box’s length in relation to a value of x is 38 cm minus
twice the value of x. |

Notice that when defining

To envision the graph of $y = V(x)$, imagine that the cutouts’ side lengths varies. The smallest that the cutouts can be is 0 cm on each side. Notice that $V(0) = 0$, because $h(0) = 0$. That is, the box’s height will be 0 when the cutout’s sides are 0 cm.

The largest that the cutouts can be is 16 cm on each side. With a side length of 16 cm the four cutouts will use all the cardboard’s width, so $w(16)$, the box’s width when $x = 16$, would be 0. Therefore $A(16) = 0$, and hence $V(16) = 0$.

As the value of

Figure 3.13.2. As the cutouts’ side lengths vary from 0 to 16, the box’s volume varies from 0 to some maximum value and then to 0.

Figure 3.13.3 shows a GC window containing definitions of

Figure 3.13.3. Function definitions for representing the volume of a box, as shown in Figure 3.24, in relation to the length of cutouts’ sides.

- Here are the function definitions from Figure 3.13.3. Explain how GC
computes $V(2.3)$. In your explanation, say what the value of each
function represents.

$V(x)=h(x)A(x)$

$h(x)=x$

$A(x)=l(x)w(x)$

$w(x)=32-2x$

$l(x)=38-2x$

$y=V(x),0<x<n$

- Explain how GC generates the animation in Figure 3.13.3 as the value
of
*n*varies. - Enter the function definitions below into GC. Click “n” on the
*n*slider and change the domain of*n*to vary from 0 to 16. Then click the play button. Explain in terms of values of*x*and*n*how GC displays the same graph as in Figure 3.13.3. - In Exercise 3.13.3, why is it essential that we use the variable
*x*in the line $y = V(x), 0 < x < n$ for GC to display the graph that it does? (Your answer should refer to the conventions built into GC.) - The animation below shows two baseball players, one on 1st base (Player 1) and one on 2nd base (Player 2). Player 1 runs at 18 ft/sec. Player 2 runs at 26 ft/sec. The distance between bases is 90 ft. The runners leave their bases simultaneously. Assume that they accelerate immediately to their running speeds. Runner 2 stops at 3rd base the moment he reaches it. Runner 1 stops at 2nd base the moment he reaches it.
- Define a function
*B*that gives the distance between Player 1 and 2nd base as a function of the number of seconds since they began running. You will need to define your function in two parts (ctrl-shift-A), one part for Player 1’s distance from 2nd base while he is running and another part for the distance after he reaches 2nd base. - Define a function
*A*that gives the distance between Player 2 and 2nd base as a function of the number of seconds since they began running. You will need to define your function in two parts (ctrl-shift-A), one part for Player 2’s distance from 2nd base while he is running and another part for the distance after he reaches 3rd base. - Define a function
*D*that gives the distance between runners as a function of the number of seconds since they began running. - Graph your function
*D*in GC. (You will need to rescale your axes to see GC's displayed graph.) Use GC’s tracing feature (click and drag on the graph) to estimate the maximum distance and the minimum distance between runners, and the number of seconds since starting at which these distances happen. If you define your functions correctly, the minimum distance will be 73.9973 feet and the maximum distance will be 94.16403 feet. - This is a challenge problem. Assume that each player accelerates to top speed at the rate of 30 (ft/sec)/sec. Rework all of a-d with this assumption.
- Enter this function into GC: $$h(x,y) = \frac{x+y}{2} - \frac{\left| x-y \right|}{2}$$
- Modify the definition of
*h*in Exercise 6 to define a function*g*that gives the maximum of*x*and*y*.

$V(u)=h(u)A(u)$

$h(s)=s$

$A(r)=l(r)w(r)$

$w(h)=32-2h$

$l(k)=38-2k$

$y=V(x),0<x<n$

$h(s)=s$

$A(r)=l(r)w(r)$

$w(h)=32-2h$

$l(k)=38-2k$

$y=V(x),0<x<n$

What does it do? How does it do what it
does?

*Hint*: $\dfrac {x + y}{2}$ gives the number half-way
between the value of *x* and the value of *y*; $\left|
x ‑ y\right|$ gives the distance between values of *x*
and *y*.