Math 3: Calculus, Muetzel
Winter 2016 - Solution 17
keywords:
extreme values, optimization
problem 1. Give an example of a function f (x) and an interval, such that f (x) does not
have a global maximum or minimum. Why does this not contradict the extreme value theorem.
Example 1
on [−1, 1].
This is not a contradiction to the extreme value theorem as f (x) is not continuous on [−1, 1].
f (x) =
1
x
Example 2
g(x) = x on (−1, 1).
This is not a contradiction to the extreme value theorem as the interval (−1, 1) is not closed.
problem 2.(10 points) 242f t of fencing are used to enclose a corral in the shape of a semicircle
attached to a rectangle. Find the diameter of the semicircle, such that the area is maximal.
We set
1.) r: radius of the semicircle.
2.) C(r) = r2 · π2 : area of the semicircle C with respect to the radius.
3.) A(r, h) = 2r · h: area of the rectangle with respect to its height h and width 2r.
4.) Constraint: The boundary B(r, h) of the corral consists of the boundary of the semicircle
and the boundary of the open rectangle. It has total length 242f t.
B(r, h) = r · π + 2r + 2h = 242 ⇔ h = 121 − (
5.) By our constraint (setting h = 0) we have that 0 ≤ r ≤
π
+ 1) · r.
2
242
π+2 .
We have to maximize A(r, h) + C(r). By our constraint we have that
A(r, h) + C(r) = r2 ·
π
π
π
π
+ 2r · h = r2 + 2r · (121 − ( + 1) · r) = 242 · r − ( + 2) · r2 .
2
2
2
2
242
242
In the interval (0, π+2
) this function has its maximum at r = π+4
' 33.9f t. This can be
conrmed by plugging in the boundary values.
Hence the area of the corral is maximal for r ' 33.9f t. The maximal value is about 4100f t2 .
Math 3: Calculus, Muetzel
Winter 2016 - Solution 17
problem 3.(10 points) (The "Catch as Catch" Can) A cylindrical can is to be made to
hold 1L of oil. Find the dimensions of the can that will minimize the metal to manufacture the
can.
This is example 2 from Stewart:Calculus, Section 3.7 on page 252. We get:
1.) r: radius of the can.
2.) A(r, h) = 2πr2 + 2πr · h: area of the can.
3.) V (r, h) = πr2 · h = 1000: volume of the can in cm3 .
Expressing h =
1000
πr2
in terms of r, we obtain:
2000
2000
and A0 (r) = 4πr − 2 .
r
r
1/3
and h = 2r. The sign test for the rst derivative
Solving for A0 (r) = 0, we obtain: r = 500
π
1/3
shows that there is indeed a minimum at r = 500
.
π
A(r) = 2πr2 +