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Basic Calculus Performance Task 2
SOLVING OPTIMIZATION PROBLEMS USING CALCULUS
- A right circular cylinder is inscribed in a sphere of radius 10 cm. Find the
dimensions of the cylinder that has maximum volume.
Important Notes:
- The formula for the volume of a right circular cylinder is V =π x
2
h, where
x
is the radius and h is the height.
- We will earmark
x (radius) as
r
cylinder
h (height) as
h
cylinder
and R (radius) as
r
sphere
for clearer understanding.
a. What is the objective? Let it be 𝑉(𝑥), where 𝑥 is the radius of the
cylinder. Use 𝑥 as the control variable.
The objective is the dimensions of the cylinder. We are required to find the
maximum volume of the right circular cylindeer inscribed in a sphere. We
let x as the radius of the cylinder, R as the radius of the sphere, and h as the
height.
b. Sketch the sphere with the inscribed cylinder. Label the important
parts.
c. Using your illustration, formulate a mathematical expression that
represents the height of the cylinder in terms of its radius.
From the figure in part b,
We can see that we can apply the Pythagorean Theorem to relate the
r
cylinder
(x) in the
h
cylinder
(h).
By Pythagorean Theorem, we have:
a
2
2
=c
2
→(r ¿¿ cylinder)
2
h
cylinder
2
=(r ¿¿ cylinder )
2
¿ ¿→(r ¿¿ cylinder)
2
h
cylinder
2
2
→(r ¿¿ cylinder)
2
h
cylinder
2
Expressing the
h
cylinder
in terms of its radius:
h
cylinder
2
= 100 −(r ¿¿ cylinder )
2
h
cylinder
100 −(r ¿¿ cylinder )
2
h
cylinder
100 −(r ¿ ¿ cylinder)
2
Now, this is the same as h= 2
100 −x
2
[since we let x (the radius of the
cylinder) as the control variable]
Therefore, the mathematical expression that represents the height of the
cylinder in terms of its radius is h= 2
100 −x
2
d. What function accurately models this problem?
As stated on the “Important Notes” number 1, we can model the function of
this problem by substituting the derived mathematical expression on letter c
in the formula V =π x
2
h.
Thus,
V =π x
2
h →=π x
2
100 −x
2
) → 2 π x
2
100 −x
2
Therefore, the function that
accurately models this problem is V ( x )= 2 π x
2
100 −x
2
e. What are the dimensions of the cylinder?
To maximize
V ( x ) , we will equate its first derivative to zero:
V ' ( x )= 0
V ' ( x )=
d
dx
( 2 π x
2
100 −x
2
→ 0 = 2 π
[
2 x
100 −x
2
2
− 2 x
2 √ 100 −x
2
]
→ 0 = 4 π
x
100 −x
2
x
3
2 √ 100 −x
2
→ 0 =x
100 −x
2
x
3
→ 0 = 2 x ( 100 −x
2
)−x
3
