WORKSHEET: Optimization problems
ANSWERS IN RED
MATH 1300
Goal: To practice solving optimization problems involving trig functions.
1. What is the area of the largest rectangle that can be inscribed in a semicircle of radius
R so that one of the sides of the rectangle lies on the diameter of the semicircle?
R
y
θ
x
First solution. Solve this problem by using the length of one side of the rectangle as a
variable. (That is, first express the area of the rectangle as a function of the variable x.)
Explain how you know that the area you found is maximal.
The area of the rectangle
√ is A = 2xy. Since the point (x, y) is on the circle of radius R,
x2 + y 2 = R2 , so y = R2 − x2 . Therefore
so
p
A = 2x R2 − x2
(0 ≤ x ≤ R),
∑
∏
p
dA
−2x
−2x2
2(R2 − x2 )
−4x2 + 2R2
= 2x √
+ 2 R2 − x2 = √
+ √
= √
.
dx
2 R2 − x2
R2 − x2
R2 − x2
R2 − x2
√
2
2
This has a critical point
in
the
interval
[0,
R]
when
−4x
+
2R
=
0,
meaning
x
=
R/
2
√
(the solution x = −R/ 2 to −4x2 + 2R2 = 0 is not in the interval [0, R]).
√
The area of the rectangle when x = R/ 2 is
√ q
√
√
√
A = 2xy = 2(R/ 2)( R2 − (R/ 2)2 ] = 2(R/ 2) · (R/ 2) = R2 .
By checking
the value of A at the endpoints x = 0 and x = R, we see that the area at
√
x = R/ 2 is indeed the absolute maximum of A on the interval [0, R].
(over)
Second solution to problem 1. Solve this problem by using the angle θ, in the drawing
on the previous page, as a variable. (That is, first express the area of the rectangle as a
function of the variable θ.)
As before, we have A = 2xy. Now from the picture on the previous page, x = R cos θ and
y = R sin θ, so
A = 2(R cos θ)(R sin θ) = 2R2 cos θ sin θ = R2 sin 2θ
by a trig identity. Then
(0 ≤ θ ≤
π
),
2
dA
= 2R cos 2θ,
dθ
which has a critical point in the interval [0, π/2] when cos 2θ = 0, so 2θ = π/2, so θ = π/4.
At this value of θ, we have
π
π
A = R2 sin 2θ = R2 sin(2( )) = R2 sin( ) = R2 ,
4
2
which is the absolute maximum by the same reasoning as on the previous page.
Looking Back. In your opinion, which approach to solving problem 1 was more straightforward (briefly explain why you hold this opinion)? The second approach, ’cause the
algebra was nicer (there weren’t any square roots involved).
2. Sue finds a tiny cigarette burn in her new rectangular scarf, which measures 5 feet by
2 feet. She decides the best solution is to cut off one corner of the scarf through the burn.
The burn is located 2 inches from one edge of the scarf and 5 inches from the end of the
scarf.
5
θ 2
{
{
h
b
(a) Show that the area of the triangle may be written as:
¥≥
¥
1≥
2 cot(θ) + 5 5 tan(θ) + 2 .
2
We have A = 12 bh; we need to relate b and h to θ. Let’s look more closely at our little
triangle in the lower right corner:
h -2
θ
θ
b -5
2
5
2
2
2
We see that tan θ = b−5
, so b − 5 = tan
θ , so b =
h−2
2 cot θ + 5. Similarly, tan θ = 5 , so b = 5 tan θ +
2. So
¥≥
¥
1
1≥
A = bh =
2 cot(θ) + 5 5 tan(θ) + 2 .
2
2
5
(over)
(b) What is the smallest area of the triangle she has to cut off to remove the burn?
(Assume the burn is a small dot and the cut is to pass through the burn.)
Multiplying out the formula from (a), we have
A=
¥≥
¥ 1≥
¥
1≥
2 cot(θ) + 5 5 tan(θ) + 2 =
10 cot θ tan θ + 4 cot θ + 25 tan θ + 10
2
2
25
= 10 + 2 cot θ +
tan θ,
2
so
so
dA
25
= −2 csc2 θ +
sec2 θ,
dθ
2
dA
dθ
= 0 when
csc2 θ
25
25
5
=
=⇒ cot2 θ =
=⇒ cot θ =
2
sec θ
4
4
2
(why not cot θ = − 52 ?) Therefore, the total area is
A = 10 + 2 cot θ +
(why is this the minimum?)
25
5 25 2
tan θ = 5 + 2 · +
· = 15
2
2
2 5