Math 7030
Activity 11- Optimization - Part II
One limitation of GSP is that one is forced to work with figures in the plane. When one
goes to solid figures, GSP is not very useful. Consider the problem of finding the
dimensions of the cylinder inscribed in a hemisphere of radius 5 (with one base of the
cylinder in the base of the hemisphere) having maximum volume.
Note that there are an infinite number of cylinders all having different shapes that fit in
the hemisphere. Consider a typical inscribed cylinder with radius r and height h. We
know that the volume V of the cylinder is V   r 2 h . However because the cylinder is
inscribed inside the hemisphere, there is a relation between h and r. It is given by the
Pythagorean Theorem
r 2  h 2  52  25
We can use this equation to solve for h in terms of r. (We could also have solved for r in
terms of h.)
h  25  r 2
Thus we can express V solely in terms of r
V   r 2 h   r 2 25  r 2
As r varies between 0 and 5 our cylinder changes and we now have a formula that gives
the volume in terms of r. We are going to analyze this formula with the aid of the TI
calculator. We first enter the formula as a function Y1
We now make a table of values of this function with the calculator and scroll down the
table. The values of r are in the X column and the values of V are in the Y1 column.
We see that V reaches a maximum somewhere around 150.8. With a step size of 0.2 in
our table, this is the best that we can do. Suppose we generate a table starting at 4 with
step size 0.05
Now we see the maximum value is around 150.89 and it occurs when the radius is
approximately 4.15.
We could continue refining our table to get more accurate results, but there is another tool
on the calculator that can be applied. First we draw a graph of the volume function. We
need to set the window. We set the X variable (which corresponds to r) to range from 0 to
5. Since the values of the volume are around 150 we set the Y range to be from 0 to 200.
We also set the Y-scale to be 10.
We then plot our graph.
The graph shows that the volume reaches a maximum a little after 4.
We press the CALC button and select the Maximum tool. The calculator prompts us for a
left bound, a right bound and an initial guess. With a left bound of 3, a right bound of 5
and an initial guess of 4 and one last press of the ENTER key, we obtain the following
screen
This indicates that the maximum volume of the cylinder is approximately 151.14955 and
it occurs when the radius is approximately 4.08248. We can then use the equation
h  25  r 2 to determine h when r = 4.08248. We get 2.88676. Thus we have a good
approximation to our original optimization problem. The key step is getting the formula
for the volume in terms of a single variable. In this case it was the Pythagorean theorem
that did the trick.
1. Use the maximum tool on the graphing calculator to find the rectangle of maximum
area inscribed in the triangle ABC as shown below.
A
3
D
E
y
x
B
F
2
C
2. The isosceles triangle ABC has a base of 3 units and a height of 4 units. An isosceles
triangle DEF with DE parallel to AB is inscribed in the triangle ABC as shown.
C
D
E
B
F
Express the area of the inscribed triangle as a function of one variable. Then use the
graphing calculator to find the area of the largest such triangle that can be inscribed in
ABC. What are the base and height of the largest such triangle?
A
3. A right circular cylinder is inscribed inside a right circular cone having a height of 10
and a base radius of 4 as shown below.
a. If the radius of the cylinder is 2, determine its volume.
b. If the radius of the cylinder is 3, determine its volume.
c. If the radius of the cylinder is r, express the volume of the cylinder in terms of the
variable r.
d. Graph the volume function of r on the graphing calculator.
e. Use the maximum tool on the calculator to determine the value of r for which the
volume is maximal. What is the maximum volume of such a cylinder?
4. A box with a square base and a square top is to hold 8000 cubic inches. The material
for the sides costs 0.25 cents per square inch while the material for the top and bottom
costs 0.75 cents per square inch.
a. If the dimensions of such a box are 20 by 20 by 20 inches, what is the cost?
b. If the dimensions of such a box are 10 by 10 by 80 inches, what is the cost?
c. If the dimension of the base is x and the height of the box is y, express the cost of
the box in terms of x and y.
d. Use the fact that the volume is 8000 to express y in terms of x (or x in terms of y).
e. Express the cost of the box as a function of a single variable.
f. Use the TI calculator to graph the cost function and then use the minimum tool on
the calculator to determine the minimum cost of such a box.
g. What are the dimensions of the cheapest box?