MATH 16020 LESSON 24 EXAMPLES
MARCH 22, 2017
Example 1. We are tasked with constructing a rectangular box with a volume of
64 cubic feet. The material for the top costs 8 dollars per square foot, the material
for the sides costs 10 dollars per square foot, and the material for the bottom costs
4 dollars per square foot. To the nearest cent, what is the minimum cost for such a
box?
Solution: Let w be the width, h be the height, and l be the length of this box.
The volume of the box is given by V = whl. The goal here is to minimize the cost
function, not the volume. In fact, we are requiring that the volume be 64 cubic feet.
We’ll call this the constraint.
By the picture above, you should see the area of the top is lw, the area of the
bottom is lw, and the total area of the sides is 2wh + 2lh (this is because we are
not assuming the box has a square base so w and l can be different). Thus, by the
information we are given above, our cost function is
C(w, h, l) = 8 (lw) +10 (2wh + 2lh) +4 (lw) = 12lw + 20wh + 20lh.
|{z}
|
{z
}
|{z}
area of
top
area of
sides
area of
bottom
Unfortunately, this is a function of 3 variables and all of our tools only work for
functions of 2 variables. We use the constraint (lwh = 64) to rewrite the cost function
as a function of only 2 variables.
Write l =
64
. Then, substituting, we get
wh
64
64
C(w, h) = 12
w + 20wh + 20
h
wh
wh
=
768
1280
+ 20wh +
.
h
w
Since this is now a function of 2 variables, we can find the critical points.
1
2
LESSON 24 EXAMPLES
Differentiating,
Cw = 20h −
1280
768
and
C
=
−
+ 20w.
h
w2
h2
Recall that the critical points are the points (w, h) that make both Cw and Ch equal
to zero. So
Cw = 0
1280
=0
w2
1280
⇒ 20h =
w2
1280
= 64
⇒ w2 h =
20
⇒ 20h −
and
Ch = 0
768
=0
h2
768
⇒ 20w = 2
h
768
192
⇒ h2 w =
=
.
20
5
⇒ 20w −
Now, we observe that
192
5
= 53 (64). This means that
h2 w =
192
3
3
= (64) = w2 h.
5
5
5
Since we are assuming the volume is 64 cubic inches, we must have w 6= 0 and h 6= 0.
So, we divide both sides by hw and our equation becomes
3
h = w.
5
Then, because w2 h = 64,
w
2
3
w
5
= 64 ⇒ w3 =
5(64)
320
=
.
3
3
Thus,
r
w=
3
320
3
and so
3
3
h= w=
5
5
r
3
320
.
3
LESSON 24 EXAMPLES
3
Our critical point is then
r
3
(w, h) =
320 3
,
3 5
r
3
320
3
!
Therefore, our cost is minimized at
!
!
r
r
r
3
768
320
320
320
3
3
3
C
,
= q
+ 20
3 5
3
3
3 3 320
5
.
3
5
r
3
320
3
!
1280
+q
3
3
320
3
≈ 809.695
Technically, you should check that this is actually a minimum by going through
the Second Derivative Test. But since this is the only critical point and it is much
easier to maximize a cost rather than minimize it, we can assume that this is indeed
a minimum.
Example 2. The post office will accept packages whose combined length and girth
is at most 50 inches (girth is the total perimeter around the package perpendicular
to the length and the length is the largest of the 3 dimensions). What is the largest
volume that can be sent in a rectangular box? (Round your answer to the nearest
integer.)
Solution: Our goal is to maximize the volume V = lwh where l is the largest of
the 3 dimensions. The girth is then g = 2w + 2h. We are told that the combined
length and girth can be at most 50 inches, to maximize the volume we will want to
maximize this as well. So we set l + g = 50.
Now, our volume function is a function of more than 2 variables so we need to
rewrite it in terms of 2. We know that l + g = 50 and that g = 2w + 2h, so
l = 50 − g = 50 − 2w − 2h.
Thus, our volume function becomes
V = lwh = (50 − 2w − 2h)wh = 50wh − 2wh2 − 2w2 h.
This is the function we want to maximize.
Step 1. Find critical points
Vw = 50h − 2h2 − 4wh which means Vw = 0 ⇒ 50h − 2h2 − 4wh = 0, so either
h = 0 or 50 − 2h − 4w = 0. Because we can’t let one of our dimensions be 0, we throw
out the case where h = 0. From 50 − 2h − 4w = 0, we get
0 = 50 − 2h − 4w
⇒ 0 = 25 − h − 2w
⇒ h = 25 − 2w
Vh = 50w − 4wh − 2w2 means Vh = 0 ⇒ 50w − 4wh − 2w2 = 0, so either w = 0
or 50 − 4h − 2w = 0. Again, we can’t let one of our dimensions be 0 so we throw out
4
LESSON 24 EXAMPLES
w = 0. So, we have 50 − 4h − 2w = 0 which means
0 = 50 − 4h − 2w
⇒ 0 = 25 − 2h − w
⇒ w = 25 − 2h
= 25 − 2 (25 − 2w)
| {z }
h
= 25 − 50 + 4w
= −25 + 4w
⇒ −3w = −25
25
.
3
⇒ w=
Thus,
h = 25 − 2w = 25 − 2
Our critical point is then (w, h) =
25
3
= 25 −
2(25)
25
= .
3
3
25 25
,
.
3 3
Step 2. Find second derivatives
Vww = −4w, Vhh = −4h, and Vwh = 50 − 4h − 4w
Step 3. Find discriminant
D = Vww Vhh − (Vwh )2 = (−4w)(−4h) − (50 − 4h − 4w)2 = 16wh − (50 − 4h − 4w)2
Step 4. Apply Test
D
25 25
,
3 3
2
25
25
25
25
= 16
− 50 − 4
−4
3
3
3
3
2 2
100
150 100 100
=
−
−
−
3
3
3
3
2 2
100
50
− −
>0
=
3
3
and
Vww
which means that
25 25
,
3 3
25 25
,
3 3
= −4
25
3
<0
is indeed a maximum.
Finally, we need to plug this into our function for volume. Since l = 50 − 2w − 2h,
25
25
150 50 50
50
l = 50 − 2
−2
=
−
−
= .
3
3
3
3
3
3
LESSON 24 EXAMPLES
5
Thus, our maximum volume is
50
25
25
31, 250
V =
=
≈ 1157.
3
3
3
27
Example 3. Suppose an ice cream factory puts ice cream into cylindrical containers.
The volume of ice cream in each container is 128π cubic inches. What is the least
amount of material required to make one container (include a simple lid for the top
of the container)? Round your answer to the nearest square inch.
Solution: We are asked to minimize the surface area of the container such that
the container’s volume is exactly 128π cubic inches. The function we want to minimize
is
SA =
2πr2
+ |2πrh
|{z}
{z }
area of
top and bottom
area of
sides
where r is the radius of the container and h is its height. To see this, consider the
picture
128
, then by substituting into
r2
SA we can actually make this a function of a single variable.
We are also given V = 128π = πr2 h. So if we let h =
Note 1. If we try to differentiate SA when it is still a function of 2 variables, the
only critical point we will find is r = 0, h = 0. That clearly doesn’t make sense, so
what’s going on? Well, technically, r = 0, h = 0 does minimize the surface area — it
makes it zero. But it doesn’t satisfy the constraint V = 128π. The only way to let
128
the function know about the constraint is by substituting h = 2 .
r
Next, write
2
2
SA = 2πr + 2πrh = 2πr + 2πr
128
r2
= 2πr2 +
256π
.
r
So we maximize SA as a function of a single variable. The method is the same: we
begin with finding the critical points. Differentiating with respect to r, we get
dSA
256π
= 4πr − 2 .
dr
r
6
LESSON 24 EXAMPLES
dSA
does
dr
not exist at r = 0, but we don’t want the volume to be 0 so we don’t consider that
dSA
= 0. Write
point. We check where
dr
Now, we need to find either where this derivative does not exist or is 0.
0 = 4πr −
⇒
256π
r2
256π
= 4πr
r2
⇒ r3 = 64
⇒ r=4
Thus, our critical point is r = 4. We want to make sure this is indeed a minimum for
our function. We check by plugging in other potential values:
256π
= 32π + 16π = 48π
4
256π
SA(1) = 2π(1)2 +
= 2π + 256π = 258π.
1
SA(4) = 2π(4)2 +
This is indeed a minimum with surface area 48π ≈ 151 square inches.
Example 4. A biologist must make a medium to grow a type of bacteria. The
percentage of salt in the medium is given by S = 0.01x2 y 2 z, where x, y, and z are
amounts in liters of 3 different nutrients mixed together to create the medium. The
ideal salt percentage for this type of bacteria is 48%. The costs of x, y, and z nutrient
solutions are respectively, 6, 3, and 8 dollars per liter. Determine the minimum cost
that can be achieved. (Round your answer to the nearest 2 decimal places.)
Solution: We want to minimize the cost function, given by
C(x, y, z) = 6x + 3y + 8z
such that S(x, y, z) = .48. We want to reduce our cost function to a function of 2
variables, which we do using S. Write
.48 = 0.01x2 y 2 z ⇒ z =
48
.
x2 y 2
Substituting this into our cost function, we get
48
384
C(x, y) = 6x + 3y + 8
= 6x + 3y + 2 2 .
2
2
xy
xy
Our next step is to go through our method.
Step 1. Find critical points
Cx = 6 −
128 = x3 y 2 .
2(384)
768
768
768
= 6 − 3 2 , so Cx = 0 implies 0 = 6 − 3 2 ⇒ 3 2 = 6 ⇒
3
2
xy
xy
xy
xy
LESSON 24 EXAMPLES
Cy = 3 −
7
768
768
768
, so Cy = 0 implies 0 = 3 − 2 3 ⇒ 2 3 = 3 ⇒ 256 = x2 y 3 .
2
3
xy
xy
xy
Now, since 256 = 2(128), we see that 2 (x3 y 2 ) = x2 y 3 ⇒ x = 0 or y = 0 or 2x = y.
| {z } |{z}
128
256
Because we want S(x, y, z) = .48 6= 0, we can’t have x = 0 or y = 0. So we have
2x = y. Then since 256 = x2 y 3 ⇒ 256 = x2 (2x)3 = 8x5 ⇒ 32 = x5 ⇒ x = 2. So
y = 2(2) = 4.
Our critical point is then (2, 4).
Step 2. Find second derivatives
2304
1536
2304
, Cyy = 2 4 , Cxy = 3 3
4
2
xy
xy
xy
Cxx =
Step 3. Find discriminant
D = Cxx Cyy − (Cxy )2
2
2304
2304
1536
=
−
x4 y 2
x2 y 4
x3 y 3
=
(2304)2 (1536)2
−
x6 y 6
x6 y 6
Step 4. Apply Test
D(2, 4) =
Cxx (2, 4) =
(2304)2
(1536)2
−
>0
(2)6 (4)6 (2)6 (4)6
2304
>0
(2)4 (4)2
Therefore, (2, 4) is a minimum.
The function value at (2, 4) is
C(2, 4) = 6(2) + 3(4) +
384
384
= 12 + 12 +
= 12 + 12 + 6 = 30.
2
2
(2) (4)
64