Applied Max and Min
Solutions To Selected Problems
Calculus 9th Edition Anton, Bivens, Davis
Matthew Staley
October 27, 2011
1. Find the dimensions of the rectangle with the maximum area that can be inscribed in a circle of radius 10.
Let x and y be as is shown in the figure above. Since the radius is 10, the
hypotenuse of the triangle is the diameter = 20. Then we have the constraint
x2 + y 2 = 202 and we must maximize area, A = xy
x2 + y 2 = 400
y 2 = 400 − x2
√
y = 400 − x2
√
A = x 400 − x2
Sub into A = xy
0 ≤ x ≤ 20
dA √
1
−2x
= 400 − x2 + x √
dx
2 400 − x2
(400 − x2 ) − x2
= √
400 − x2
2(200 − x2 )
= √
=0
400 − x2
1
→ 200 − x2 = 0
x2 = 200
√
√
x = 200 = 10 2
→ y 2 = 400 − 200
y 2 = 200
√
√
y = 200 = 10 2
√
The dimensions x = y = 10 2 show that a square has the maximum area
inscribed in a circle.
2. A box with a square base is taller than it is wide. In order to send the box
through the U.S. mail, the height of the box and the perimeter of the base can
sum to no more than 108 inches. What is the maximum volume for such a
box?
Let x and y be as is shown in the figure above. The perimeter of the base is
then 4x and the height is y. We have the constraint equation 4x + y ≤ 108 and
2
we want to maximize the volume V = x2 y.
Let 4x + y = 108
y = 108 − 4x
V = x2 y = x2 (108 − 4x)
= 108x2 − 4x3
dV
= (2)180x − (3)4x2
dx
= 4x(54 − 3x) = 0
→ 4x = 0
x=0
→ 54 − 3x = 0
3x = 54
x = 18
Take x = 18 as x = 0 is not allowed for our dimension. Substitute x into the
equation y = 108 − 4x to find y:
y = 108 − 4(18)
= 108 − 72
= 36
Therefore our maximum volume will be
V = x2 y
= (18)2 (36)
= 324(36)
= 11664
3
3. Find the dimensions of the right circular cylinder of largest volume that can
be inscribed in a sphere of radius R.
Let r and h be the dimensions as shown in the figure above. The volume of
the of inscribed cylinder is V = πr2 h.
2
h
2
r +
= R2
2
h2
r 2 = R2 −
4
h2
h
→V =π R −
4
h3
2
= π hR −
4
2
0 ≤ h ≤ 2R
dV
3 2
2
=π R − h =0
dh
4
4
3
→ R2 − h2 = 0
4
4
h2 = R2
3
r
4 2
2R
h=
R = √
3
3
h2
4
4
R2
= R2 −
3 4
1
= R2 − R2
3
2 2
= R
3
r
2
r=
R
3
→ r 2 = R2 −
4. A cylindrical can, open on top, is to hold 500 cm3 of liquid. Find the height
and radius that minimizes the amount of material needed to manufacture the
can.
5
The amount of material needed correlates to the surface area of the can. The
surface area of the bottom is just the area of the circle πr2 . Since there is an
open top, there is no surface area there. The remaining surface area is 2πhr.
We have our surface area equation S = πr2 +2πrh, which we want to minimize.
500
Our constraint is that V = πr2 h = 500 → h = πr
2
500
2
S = πr + 2πr
πr2
1000
= πr2 +
r>0
r
dS
1000
2πr3 − 1000
= 2πr − 2 =
=0
dr
r
r2
→ 2πr3 = 1000
500
r3 =
π
r
3 500
r=
π
500
500 π 2/3
→h= 2 =
=
πr
π 500
6
r
3
500
π
5. A cone shaped paper drinking cup is to hold 100 cm3 of water. Find the height
and radius of the cup that will require the least amount of paper. Let r, h, L
be √as shown in the figure above. The area of the paper is then A = πrL =
πr r2 + h2 since L2 = r2 + h2 . We also have the constraint V = 31 πr2 h =
300
100 → h = πr
2 . Sub this into A:
s
A = πr
r2 +
300
πr2
2
r
= πr
r2 +
90000
π 2 r4
To simplify computations, let S = A2 .
2 2
→S=π r
90000
r + 2 4
π r
2
= π 2 r4 +
90000
r2
dS
180000
4π 2 r6 − 180000
= 4π 2 r3 −
=
=0
dr
r3
r3
7
→ 4π 2 r6 − 180000 = 0
4π 2 r6 = 180000
180000
45000
r6 =
=
2
4π
π2
r
→r=
→h=
6
45000
π2
300
πr2
300
=
π
π2
45000
2/6
300
=
π
π2
45000
1/3
300
h=
π
r
3
π2
45000
8
6. A trapezoid is inscribed in a semicircle of radius 2 so that one side is along the
diameter. Find the maximum possible are for the trapezoid.
By the figure above the area of the trapezoid is :
1
A = (b1 + b2 )h
2
1
= (4 + 4 cos(θ))(sin(θ))
2
= 4(sin(θ) + sin(θ) cos(θ))
0 ≤ θ ≤ π/2
dA
= 4(cos(θ) − sin2 (θ) + cos2 (θ))
dθ
= 4(cos(theta) − [1 − cos2 (θ)] + cos2 (θ))
= 4(2 cos2 (θ) + cos(θ) − ‘1)
= 4(2 cos(θ) − 1)(cos(θ) + 1)
= 0 when θ = π/3 f or 0 < θ < π/2
9
So the maximum area occurs when θ = π/3 which gives :
A = 4(sin(θ) + sin(θ) cos(θ))
= 4(sin(π/3) + sin(π/3) cos(π/3))
√
√
= 4( 3/2 + 3/2(1/2))
√
√
=2 3+ 3
√
= 3 3
10