Calculus I (MAC2311-v3)
Quiz 5 (2014/11/04)
Name (PRINT):
An answer with no work or reasoning receives no credit.
You may not use any calculators.
1. (7 points) Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r. (Sketch a diagram with all the necessary quantities clearly
labeled.)
Solution:
Method (1): Express area in terms of y using geometry; maximize using
calculus.
The area of the triangle is given by x(r + y). We can write x in terms of y and r
r 2 = x2 + y 2 ,
p
x = r2 − y 2 .
Now, the area can be written as
A1 (y) =
p
r2 − y 2 (r + y).
Geometrically, A1 (y) is defined only when y is in D1 = [−r, r]. (A negative value
for y corresponds to the base of the triangle being above the center of the circle.
The picture shows the case where the base is below the center, corresponding to a
positive value for y.) At the end points y = −r or y = r the triangle degenerates
into a point or line.
MAC2311
Page 1 of 6
Differentiating we get
p
−2y
A01 (y) = p
(r + y) + r2 − y 2
2 r2 − y 2
p
y 2 + ry
= r2 − y 2 − p
.
r2 − y 2
Setting A01 (y) to zero gives
p
y 2 + ry
2
2
p
r −y =
,
r2 − y 2
r2 − y 2 = y 2 + ry,
2y 2 + ry − r2 = 0,
(2y − r)(y + r) = 0,
leading to the critical points y = r/2 and y = −r. Other critical points come from
the points y at which A01 (y) is not defined:
r2 − y 2 = 0,
giving the new critical point y = r. Evaluting A1 at the end points of the domain
D1 and at the critical points we get
A1 (−r) = 0,
A1 (r) = 0,
r
r
r
r
A1 ( ) = r2 − ( )2 (r + ),
2
2
2
√ 2
3 3r
=
.
4
So, the dimensions of the isosceles triangle of largest area inscribed in a circle of
radius r are
r
√
r
base = 2x = 2 r2 − ( )2 = 3r,
2
3r
height = r + y = .
2
In other words, the optimal isosceles triangle (in the sense above) is an equilateral
triangle. (This may have been expected from symmetry.)
Method (2): Express area in terms of x using geometry; maximize using
calculus.
MAC2311
Page 2 of 6
The area of the triangle is given by x(r + y). We can write y in terms of x and r
r 2 = x2 + y 2 ,
√
y = r 2 − x2 .
Now, the area can be written as
A2 (x) = x(r +
√
r2 − x2 ).
Geometrically, A2 (x) is defined only when x is in D2 = [0, r]. At the end point x = 0
the triangle degenerates into a line.
Differentiating we get
√
−2x
A02 (x) = (r + r2 − x2 ) + x( √
)
2 r 2 − x2
√
r r 2 − x2 + r 2 − x2 − x 2
√
=
r 2 − x2
√
r r2 − x2 + r2 − 2x2
√
=
.
r 2 − x2
Setting A02 (x) to zero gives
√
r r2 − x2 = 2x2 − r2 ,
(1)
which upon squaring both sides (which may create extra solutions; we need to check)
leads to
r2 (r2 − x2 ) = 4x4 − 4r2 x2 + r4 ;
MAC2311
(2)
Page 3 of 6
that is,
x2 (4x2 − 3r2 ) = 0.
Of the two solutions
(3)
√
3r
2
of this equation, x = 0 does not satisfy
the original equation (1) and must be
√
3r
discarded. So, one critical point is x = 2 . Setting the denominator of A02 (x) to
zero gives r2 − x2 = 0, leading to the other critical point x = r. Evaluting A2 at the
end points of D2 and at the critical points we get
x = 0 and x =
A2 (0) = 0,
A2 (r) = r2 ,
r
√
√
3r
3r
3r2
A2 (
)=
(r + r2 −
)
2
2
4
√
3 3r2
=
.
4
So, the dimensions of the isosceles triangle of largest area inscribed in a circle of
radius r are
√
3r √
base = 2x = 2
= 3r,
2 s
√
3r 2 3r
height = r + y = r + r2 − (
) = .
2
2
Method (3): Express area in terms of θ using trigonometry; maximize
using calculus.
MAC2311
Page 4 of 6
The area of the triangle is given by (height)(base/2); that is,
height = (side) cos(θ) = 2r cos(θ) cos(θ) = 2r cos2 (θ),
base/2 = (side) sin(θ) = 2r cos(θ) sin(θ),
A3 (θ) = 2r cos2 (θ)2r cos(θ) sin(θ)
= 4r2 sin(θ) cos3 (θ).
Geometrically, A3 (θ) is defined only when θ is in D3 = [0, π/2]. At the end points
θ = 0 or θ = π/2 the triangle degenerates into a line or a point.
Differentiating we get
A03 (θ) = 4r2 cos(θ) cos3 (θ) + sin(θ)(3 cos2 (θ))(− sin(θ))
= 4r2 cos2 (θ) cos2 (θ) − 3 sin2 (θ) .
Setting A3 (θ) to zero gives
cos2 (θ) = 0,
cos2 (θ) − 3 sin2 (θ) = 0,
the latter giving
1
tan2 (θ) = ,
3
1
tan(θ) = √ .
3
In our domain D3 , the first equation gives θ = π/2, while the second gives θ = π/6.
These are our critical points.
Evaluating A3 at these critical points and the end points of our domain we get
A3 (0) = 0,
A3 (π/2) = 0,
A3 (π/6) = 4r2 sin(π/6) cos3 (π/6)
√
= 4r2 (1/2)( 3/2)3
√
3 3r2
=
.
4
So, the dimensions of the isosceles triangle of largest area inscribed in a circle of
radius r are
√
height = 2r cos2 (θ) = 2r cos2 (π/6) = 2r( 3/2)2
3r
= ,
2
√
base = 4r cos(θ) sin(θ) = 4r(cos(π/6)) sin(π/6) = 4r( 3/2)(1/2)
√
= 3r.
MAC2311
Page 5 of 6
2. (3 points) Using Newton’s method and starting from the initial approximation of x1 = 0,
find x2 , the second approximation to the root of the equation x3 − e−x = 0.
Solution: Put f (x) = x3 − e−x . Differentiating we get f 0 (x) = 3x2 + e−x . At x1 = 0
we have f (x1 ) = −1 and f 0 (x1 ) = 1.
Starting Newton’s iteration
xn+1 = xn −
f (xn )
f 0 (xn )
from x1 = 0 we get
f (x1 )
f 0 (x1 )
−1
=0−
1
= 1.
x2 = x1 −
MAC2311
Page 6 of 6