Suppose you want to find the maximum value of a surface z=f(x,y) on the boundary of the unit circle, x2+y2=1.
True or false: you should always check the points (0,±1) and (±1,0), since these are the endpoints of the circle.
Hint+
What is an endpoint of a circle?
Answer+
false
Full solution+
False. A common mistake is to think that the intercepts of a circle are somehow “endpoints," in the same way that the interval [−1,1] has endpoints −1 and 1. But circles don't have endpoints!
When (At least in this section, this is how we do it.. but we'll learn other ways that also don't involve optimizing x and y separately) we're finding the extrema of a function over a closed curve, we use the equation of the curve
to get a function of one variable. Then we look for critical points and endpoints of that function. These may or may not occur at x=±1 or y=±1.
Now, you might notice that school problems often end up having their extrema at the extreme values of x and/or y in the boundary. This is a result of writing problems with relatively easy algebra, rather than the result of some universal law.
Find the high and low points of the surface z=x2+y2 with (x,y)
varying over the square ∣x∣≤1, ∣y∣≤1. Discuss the values of
zx,zy there. Do not evaluate any derivatives in answering this
question.
Hint+
Interpret the height x2+y2 geometrically.
Answer+
The minimum height is zero at (0,0,0). The derivatives
zx and zy do not exist there.
The maximum height is 2 at (±1,±1,2).
There zx and zy exist but are not zero — those points would
not be the highest points if it were not for the restriction
∣x∣,∣y∣≤1.
Full solution+
The height x2+y2 at (x,y) is the distance from (x,y)
to (0,0). So the minimum height is zero at (0,0,0). The surface is a
cone. The cone has a point at (0,0,0) and the derivatives
zx and zy do not exist there.
The maximum height is achieved when (x,y) is as far as possible from
(0,0). The highest points are at (±1,±1,2). There
zx and zy exist but are not zero. These points would not
be the highest points if it were not for the restriction ∣x∣,∣y∣≤1.
The specified function and its first order derivatives are
f(x,y)=xy−x3y2fx(x,y)=y−3x2y2fy(x,y)=x−2x3y
First, we find the critical points. The first partial derivatives are defined everywhere and so the critical points are the solution of
fxfy=0=0⟺⟺y(1−3x2y)x(1−2x2y)=0=0⟺⟺y=0 or 3x2y=1x=0 or 2x2y=1
If y=0, we cannot have 2x2y=1, so we must have x=0.
If 3x2y=1, we cannot have x=0, so we must have 2x2y=1. Dividing
gives 1=2x2y3x2y=23 which is impossible.
So the only critical point in the square is (0,0). There f=0.
Next, we look at the part of the boundary with x=0. There f=0.
Next, we look at the part of the boundary with y=0. There f=0.
Next, we look at the part of the boundary with x=1. There f=y−y2.
As dyd(y−y2)=1−2y, the max and min of y−y2 for
0≤y≤1 must occur either at y=0, where f=0, or at y=21,
where f=41, or at y=1, where f=0.
Next, we look at the part of the boundary with y=1. There f=x−x3.
As dxd(x−x3)=1−3x2, the max and min of x−x3 for
0≤x≤1 must occur either at x=0, where f=0, or
at x=31, where f=332,
or at x=1, where f=0.
All together, we have the following candidates for max and min.
point
(0,0)
x=0
y=0
(1,0)
(1,21)
(1,1)
(0,1)
(31,1)
(1,1)
value of f
0
0
0
0
41
0
0
332
0
min
min
min
min
min
min
max
min
The largest and smallest values of f in this table are
Find and classify the critical points of h(x,y) as local maxima, local minima or
saddle points.
Find the maximum and minimum values of h(x,y) on the disk x2+y2≤1.
Answer+
(a)
pointcritical
type
(0,32)
local max
(0,−32)
local min
(2,0)
saddle point
(−2,0)
saddle point
(b) The maximum and minimum values of h(x,y) in x2+y2≤1
are 3 (at (0,1)) and −3 (at (0,−1)), respectively.
Full solution+
(a)
To find the critical points we will need the
first order partial derivatives of h and to apply the second derivative test of
Theorem 16.1.14 in the text
we will need all
second order partial derivatives. So we need all partial derivatives of
f up to order two.
Here they are.
(b) The absolute max and min can occur either in the interior
of the disk or on the boundary of the disk. The boundary
of the disk is the circle x2+y2=1.
Any absolute max or min in the interior of the disk
must also be a local max or min and, must also be a critical point of h.
We found all of the critical points of h in part (a).
Since 2>1 and 32>1 none of the critical
points are in the disk.
At each point of x2+y2=1 we have
h(x,y)=3y
with −1≤y≤1. Clearly the maximum value is 3
(at (0,1)) and the minimum value is −3 (at (0,−1)).
So all together, the maximum and minimum values of h(x,y) in x2+y2≤1
are 3 (at (0,1)) and −3 (at (0,−1)), respectively.
Find the absolute maximum and minimum
values of the function
f(x,y)=5+2x−x2−4y2 on the rectangular region
R={(x,y)−1≤x≤3,−1≤y≤1}
Hint+
There are five places to check: the interior and four boundaries.
Answer+
The minimum is −2 and the maximum is 6.
Full solution+
The maximum and minimum must either occur at a critical point or on
the boundary of R.
The critical points are the points where the first order partial derivatives are zero or one does not exist. Here fx(x,y)=2−2x and fy(x,y)=−8y and so they are defined everywhere. Therefore, the critical points are the solutions of
00=fx(x,y)=2−2x=fy(x,y)=−8y
So the only critical point is (1,0).
On the side x=−1, −1≤y≤1 of the boundary of R
f(−1,y)=2−4y2
This function decreases as ∣y∣ increases. So its maximum value
on −1≤y≤1 is achieved at y=0 and its minimum value is achieved
at y=±1.
On the side x=3, −1≤y≤1 of the boundary of R
f(3,y)=2−4y2
This function decreases as ∣y∣ increases. So its maximum value
on −1≤y≤1 is achieved at y=0 and its minimum value is achieved
at y=±1.
On both sides y=±1, −1≤x≤3 of the boundary of R
f(x,±1)=1+2x−x2=2−(x−1)2
This function decreases as ∣x−1∣ increases. So its maximum value
on −1≤x≤3 is achieved at x=1 and its minimum value is achieved
at x=3 and x=−1 (both of whom are a distance 2 from x=1).
So we have the following candidates for the locations of the min and max
Find the minimum of the function h(x,y)=−4x−2y+6 on the closed
bounded domain defined by x2+y2≤1.
Answer+
6−25
Full solution+
Since ∇h=⟨−4,−2⟩ exists and is never zero,
h has no critical points and the minimum of h on the disk
x2+y2≤1 must be taken on the boundary, x2+y2=1, of the
disk.
To find the minimum on the boundary, we need to use the equation x2+y2≤1
to turn h(x,y) into a function of one variable. We can break the boundary up into two pieces: y=1−x2, −1≤x≤1, and
y=−1−x2, −1≤x≤1.
Define g1(x) as the value of h along the boundary curve y=1−x2, −1≤x≤1.
All together, the minimum value h achieves over the boundary x2+y2=1 is
6−25. Since we already decided the global minimum would occur on the boundary, that tells us our global minimum is 6−25.
Find all critical points of f, and classify each one as a local maximum,
a local minimum, or saddle point.
Find the location and value of the absolute maximum and minimum of f
on the triangular region x≥0, y≥0, x+y≤8.
Hint+
Since the region is a triangle, your boundary will have three separate parts to check.
Answer+
(a)
(0,0) and (3,0) and (0,3) are saddle points
(1,1) is a local min
(b)
The minimum is −1 at (1,1) and the maximum is 80 at (4,4).
Full solution+
(a)
Thinking a little way ahead, to find the critical points we will need the
first order partial derivatives of f and to apply the second derivative test of
Theorem 16.1.14 in the text
we will need all
second order partial derivatives. So we need all partial derivatives of
f up to order two.
Here they are.
(Of course, fxy and fyx have to be the same. It is still
useful to compute both, as a way to catch some mechanical errors.)
The first order partial derivatives are defined everywhere and so the critical points are the solutions of
fx=y(2x+y−3)=0fy=x(x+2y−3)=0
The first equation is satisfied if at least one of y=0, y=3−2x
are satisfied.
If y=0, the second equation reduces to x(x−3)=0, which is
satisfied if either x=0 or x=3.
If y=3−2x, the second equation reduces to x(x+6−4x−3)=x(3−3x)=0
which is satisfied if x=0 or x=1.
So there are four critical points: (0,0), (3,0), (0,3),
(1,1).
The classification is
pointcritical
fxxfyy−fxy2
fxx
type
(0,0)
0×0−(−3)2<0
saddle point
(3,0)
0×6−(3)2<0
saddle point
(0,3)
6×0−(3)2<0
saddle point
(1,1)
2×2−(1)2>0
2
local min
(b) The absolute max and min can occur either in the interior
of the triangle or on the boundary of the triangle. The boundary
of the triangle consists of the three line segments.
Any absolute max or min in the interior of the triangle
must also be a local max or min and, must also be a critical point of f.
We found all of the critical points of f in part (a).
Only one of them, namely (1,1) is in the interior of the
triangle. (The other three critical points are all on the
boundary of the triangle.) We have f(1,1)=−1.
At each point of L1 we have x=0 and so f(x,y)=0.
At each point of L2 we have y=0 and so f(x,y)=0.
At each point of L3 we have
f(x,y)=x(8−x)(5)=40x−5x2=5[8x−x2]
with 0≤x≤8. As
dxd(40x−5x2)=40−10x, the max and min of
40x−5x2 on 0≤x≤8 must be one of
5[8x−x2]x=0=0 or
5[8x−x2]x=8=0 or
5[8x−x2]x=4=80.
So all together, we have the following candidates for max and min, with the
max and min indicated.
the first order partial derivatives are defined everywhere and the critical points are the solutions of
⟺⟺⟺fx=0y=x2y=x2y=x2fy=0y−3x+2=0x2−3x+2=0x=1 or 2
So, there are two critical points: (1,1),(2,4).
pointcritical
fxxfyy−fxy2
fxx
type
(1,1)
12×2−(−6)2<0
saddle point
(2,4)
24×2−(−6)2>0
24
local min
(b)
There are no critical points in the interior of the allowed region,
so both the maximum and the minimum occur only on the boundary. The boundary
consists of the line segments (i) x=1, 0≤y≤1, (ii) y=1, $0\le
x\le 1$ and (iii) y=1−x, 0≤x≤1.
First, we look at the part of the boundary with x=1. There f=y2−2y+2.
As dyd(y2−2y+2)=2y−2 vanishes only at y=1,
the max and min of y2−2y+2 for 0≤y≤1 must occur either at y=0,
where f=2, or at y=1, where f=1.
Next, we look at the part of the boundary with y=1. There f=2x3−6x+5.
As dxd(2x3−6x+5)=6x2−6,
the max and min of 2x3−6x+5 for 0≤x≤1 must occur either at x=0,
where f=5, or at x=1, where f=1.
Next, we look at the part of the boundary with y=1−x. There $f=2x^3-6x(1-x)
+(1-x)^2+4(1-x)=2x^3+7x^2-12x+5$.
As $\diff{}{x}(2x^3+7x^2-12x+5)=6x^2+14x-12
=2\big(3x^2+7x-6\big)
=2(3x-2)(x+3)$,
the max and min of 2x3+7x2−12x+5 for 0≤x≤1 must occur either at x=0,
where f=5, or at x=1, where f=2, or at x=32, where $f=2(\frac{8}{27})-6(\frac{2}{3})(\frac{1}{3})+\frac{1}{9}+\frac{4}{3}
=\frac{16-36+3+36}{27}
=\frac{19}{27}$.
So all together, we have the following candidates for max and min, with the
max and min indicated.
The first partial derivatives are defined everywhere. So the critical points are the solution of
fx(x,y)=fy(x,y)=0⟺2y(x+y−3)=0 and x(x+4y−6)=0⟺{y=0 or x+y=3} and {x=0 or x+4y=6}⟺{x=y=0} or {y=0,x+4y=6} or {x+y=3,x=0} or {x+y=3,x+4y=6}⟺(x,y)=(0,0) or (6,0) or (0,3) or (2,1)
Here is a table giving the classification of each of the four critical
points.
pointcritical
fxxfyy−fxy2
fxx
type
(0,0)
0×0−(−6)2<0
saddle point
(6,0)
0×24−62<0
saddle point
(0,3)
6×0−62<0
saddle point
(2,1)
2×8−22>0
2
local min
(b) Observe that xy=4 and x+2y=6 intersect when x=6−2y and
(6−2y)y=4⟺2y2−6y+4=0⟺2(y−1)(y−2)=0⟺(x,y)=(4,1) or (2,2)
The shaded region in the sketch below is D.
None of the critical points are in D. So the max and min must occur
at either (2,2) or (4,1) or on xy=4, 2<x<4 (in which case
F(x)=f(x,x4)=4(x+x8−6) obeys
F′(x)=4−x232=0⟺x=±22) or on
x+2y=6, 2<x<4 (in which case f(x,y) is identically zero). So the
min and max must occur at one of
(x,y)
f(x,y)
(2,2)
2×2(2+2×2−6)=0
(4,1)
4×1(4+2×1−6)=0
(22,2)
4(22+22−6)<0
The maximum value is 0 and the minimum value is $4(4\sqrt{2}-6)
\approx -1.37$.
A metal plate is in the form of a semi-circular disc bounded
by the x-axis and the upper half of x2+y2=4. The temperature at
the point (x,y) is given by
T(x,y)=ln(1+x2+y2)−y.
Find the
coldest point on the plate, explaining your steps carefully. (Note:
ln2≈0.693, ln5≈1.609)
Hint+
Plugging in the boundaries should be quite easy if you choose your variables wisely
Answer+
The coldest temperture is −0.391 and the coldest point is (0,2).
Full solution+
The coldest point must be either on the boundary of the plate
or in the interior of the plate.
On the semi–circular part of the boundary 0≤y≤2 and
x2+y2=4 so that T=ln(1+x2+y2)−y=ln5−y. The smallest
value of ln5−y is taken when y is as large as possible,
i.e. when y=2, and is ln5−2≈−0.391.
On the flat part of the boundary, y=0 and
−2≤x≤2 so that T=ln(1+x2+y2)−y=ln(1+x2).
The smallest value of ln(1+x2) is taken when x is as small as possible, i.e. when x=0, and is 0.
If the coldest point is in the interior of the plate,
it must be at a critical point of T(x,y). Since
Tx(x,y)=1+x2+y22xTy(x,y)=1+x2+y22y−1
a critical point must have x=0 and 1+x2+y22y−1=0,
which is the case if and only if x=0 and 2y−1−y2=0. So the only
critical point is x=0,y=1, where T=ln2−1≈−0.307.
Since −0.391<−0.307<0, the coldest temperature is −0.391 and the
coldest point is (0,2).
The first partial derivatives are defined everywhere. So the critical points are the solution of
gx(x,y)=gy(x,y)=0⟺2x=0 and −10−2y=0⟺(x,y)=(0,−5)
Since gxx(0,−5)gyy(0,−5)−gxy(0,−5)2=2×(−2)−02<0,
the critical point is a saddle point.
(b)
The extrema must be either on the boundary of the region
or in the interior of the region.
On the semi-elliptical part of the boundary −2≤y≤0
and x2+4y2=16 so that g=x2−10y−y2=16−10y−5y2=21−5(y+1)2. This has
a minimum value of 16 (at y=0,−2) and a maximum value of 21 (at y=−1).
You could also come to this conclusion by checking the critical point
of 16−10y−5y2 (i.e. solving dyd(16−10y−5y2)=0)
and checking the end points of the allowed interval (namely y=0 and y=−2).
On the flat part of the boundary y=0 and
−4≤x≤4 so that g=x2.
The smallest value is taken when x=0 and is 0 and the largest value
is taken when x=±4 and is 16.
If an extremum is in the interior of the plate,
it must be at a critical point of g(x,y). The only critical point is
not in the prescribed region.
Here is a table giving all candidates for extrema:
(x,y)
g(x,y)
(0,−2)
16
(±4,0)
16
(±12,−1)
21
(0,0)
0
From the table the smallest value of g is 0 at (0,0)
and the largest value is 21 at (±23,−1).
Equal-angle bends are made at equal distances from the two ends of a
100 metre long fence, so that the resulting three-segment fence can be placed
along an existing wall to make an enclosure of trapezoidal shape. What
is the largest possible area for such an enclosure?
Hint+
Suppose that the bends are made a distance x from the ends of the fence
and that the bends are through an angle θ. Draw a sketch of the
enclosure and figure out its area, as a function of x and θ.
Answer+
32500
Full solution+
Suppose that the bends are made a distance x from the ends of the fence
and that the bends are through an angle θ.
Here is a sketch of the enclosure.
It consists of a rectangle, with side lengths 100−2x and
xsinθ, together with two triangles, each of height xsinθ
and base length xcosθ. So the enclosure has area
Note that Ax and Aθ are define everywhere in their domain and so to find the critical points we only needed to find the points where the first order partial derivatives are zero.
Here we have used that the fence of maximum area cannot have sinθ=0
or x=0, because in either of these two cases, the area enclosed will be zero.
The first equation forces cosθ=−2x100−4x
and hence cos(2θ)=2cos2θ−1=2x2(100−4x)2−1.
Substituting these into the second equation gives
Find the most economical shape of a rectangular box that has
a fixed volume V and that has no top.
Hint+
Suppose that the box has side lengths x, y and z.
Answer+
The box has dimensions (2V)1/3×(2V)1/3×2−2/3V1/3.
Full solution+
Suppose that the box has side lengths x, y and z.
Here is a sketch.
Because the box has to have volume V we need that V=xyz.
We wish to minimize the area A=xy+2yz+2xz of the four sides and bottom.
Substituting in z=xyV,
AAxAy=xy+2xV+2yV=y−2x2V=x−2y2V
To minimize, we want Ax=Ay=0, which is the case when yx2=2V,xy2=2V.
This forces yx2=xy2. Since V=xyz is nonzero, neither x nor y
may be zero. So x=y=(2V)1/3, z=2−2/3V1/3.
The temperature T(x,y) at a point of the xy–plane is given by
T(x,y)=20−4x2−y2
Find the maximum and minimum values of T(x,y) on the disk D
defined by x2+y2≤4.
Suppose the ant is constrained to stay on the curve y=2−x2.
Where should the ant go if it wants to be as warm as possible?
Answer+
(a) The maximum and minimum values of T(x,y) in x2+y2≤4
are 20 (at (0,0)) and 4 (at (±2,0)), respectively.
(b) (0,2)
Full solution+
(a)
The maximum and minimum can occur either in the interior
of the disk or on the boundary of the disk. The boundary
of the disk is the circle x2+y2=4.
Any absolute max or min in the interior of the disk
must also be a local max or min and must also be a critical point of h.
Since Tx=−8x and Ty=−2y, the only critical point is
(x,y)=(0,0), where T=20. Since 4x2+y2≥0,
we have T(x,y)=20−4x2−y2≤20. So the maximum value
of T (even in R2) is 20.
At each point of x2+y2=4 we have
T(x,y)=20−4x2−y2=20−4x2−(4−x2)=16−3x2
with −2≤x≤2. So T is a minimum when x2 is a maximum.
Thus the minimum value of T on the disk is 16−3(±2)2=4.
So all together, the maximum and minimum values of T(x,y) in x2+y2≤4
are 20 (at (0,0)) and 4 (at (±2,0)), respectively.
(b) We are being asked to find the (x,y)=(x,2−x2) which maximizes
T(x,2−x2)=20−4x2−(2−x2)2=16−x4
The maximum of 16−x4 is obviously 16 at x=0. So the ant should
go to (0,2−02)=(0,2).
Find the largest and smallest values of x2y2z in the part of the
plane 2x+y+z=5 where x≥0, y≥0 and z≥0.
Also find all points where those extreme values occur.
Hint+
If (x,y,z) is on the plane, then you know z=5−2x−y. So, you can write x2y2z as a function of only x and y by eliminating z.
Answer+
The minimum value is 0 on
{(x,y,z)x≥0,y≥0,z≥0,2x+y+z=5,at least one of x,y,z zero}
The maximum value is 4 at (1,2,1).
Full solution+
The region of interest is
D={(x,y,z)x≥0,y≥0,z≥0,2x+y+z=5}
First observe that, on the boundary of this region, at least one of
x, y and z is zero. So f(x,y,z)=x2y2z is zero on the boundary.
As f takes values which are strictly bigger than zero at all points of D
that are not on the boundary, the minimum value of f is 0 on
∂D={(x,y,z)x≥0,y≥0,z≥0,2x+y+z=5,at least one of x,y,z zero}
The maximum value of f will be taken at a critical point.
On D
(note that the first order partial derivatives are defined everywhere) or, dividing by the first equation by xy2 and the second equation
by x2y, (recall that x,y=0)
10−6x−2y10−4x−3y=0=0or3x+yor4x+3y=5=10
Substituting y=5−3x, from the first equation, into the second equation
gives
4x+3(5−3x)=10⟹−5x+15=10⟹x=1,y=5−3(1)=2
So the maximum value of f is (1)2(2)2(5−2−2)=4 at (1,2,1).
Show that the function
f(x,y)=2x+4y+xy1
has exactly one critical point in the first quadrant x>0, y>0, and
find its value at that point.
Use the second derivative test to classify the critical point in part (a).
Explain why the inequality 2x+4y+xy1≥6
is valid for all positive real numbers x and y.
Answer+
(a) x=1, y=21, f(1,21)=6 (b) local minimum
(c) As x or y tends to infinity (with the other at least zero),
2x+4y tends to +∞. As (x,y) tends to any point on the first quadrant part of the x- and y–axes, xy1 tends to +∞.
Hence as x or y tends to the boundary of the first quadrant
(counting infinity as part of the boundary), f(x,y) tends to +∞.
As a result (1,21) is a global (and not just local) minimum
for f in the first quadrant. Hence f(x,y)≥f(1,21)=6
for all x,y>0.
Full solution+
(a) For x,y>0, fx and fy are well-defined and so the critical points are the solutions of
fxfy=2−x2y1=0⟺y=2x21=4−xy21=0
Substituting y=2x21, from the first equation, into the second gives
4−4x3=0
which forces x=1, y=21. At x=1, y=21,
f(1,21)=2+2+2=6
(b) The second derivatives are
fxx(x,y)=x3y2fxy(x,y)=x2y21fyy(x,y)=xy32
In particular
fxx(1,21)=4fxy(1,21)=4fyy(1,21)=16
Since $f_{xx}\big(1,\half\big)f_{yy}\big(1,\half\big)-
f_{xy}\big(1,\half\big)^2=4\times 16-4^2=48>0$ and fxx(1,21)=4>0,
the point (1,21) is a local minimum.
(c) As x or y tends to infinity (with the other at least zero),
2x+4y tends to +∞. As (x,y) tends to any point on the first quadrant part of the x- and y–axes, xy1 tends to +∞.
Hence as x or y tends to the boundary of the first quadrant
(counting infinity as part of the boundary), f(x,y) tends to +∞.
As a result (1,21) is a global (and not just local) minimum
for f in the first quadrant. Hence f(x,y)≥f(1,21)=6
for all x,y>0.
Let a be a constant real number. Find all points on the surface
z=f(x,y)=x2+y2
that have minimum distance from the point (0,0,a).
Hint+
The answer will be piecewise, depending on what exactly a is.
Answer+
If a<21, then the closest point is the origin. If a≥21, then the closest points are the level curve where z=a−21.
Full solution+
First, let's visualize what's going on. Our surface looks like a bowl, sitting on the origin, opening upwards. It is radially symmetric about the z-axis, with circular level curves. That means every point on a level curve is equidistant from the z-axis. Since the point (0,0,a) is on the z-axis, if there is a point (x,y,z) that has minimum distance to the point, then its entire level curve has the same minimum distance. So we expect our answer to look like a circle (or possibly a single point – a “circle" of radius 0). If a is a negative number, it seems natural that the closest point would be (0,0,0).
The distance from (0,0,a) to an arbitrary point (x,y,z) is x2+y2+(z−a)2. If the point (x,y,z) is on our surface, then z=x2+y2. Rather than deal with square roots, we'll minimize the distance squared:
f(x,y)=x2+y2+(x2+y2−a)2
From our observations above, there will be no global maximum; the global minimum will be a local minimum; the global minimum will depend on a in a less-than-simple way; and there are likely to be multiple points that are all minimum distance to (0,0,a).
The Scranton branch of a well-known paper company has two sizes of paper for sale - A4 and A3.
Each ream of A4 is sold at $6; each ream of A3 is sold at $8. Assume that every ream produced is sold.
Suppose x is the quantity of materials that go into making A4 and y is the quantity of materials that go into making A3. Then the costs involved in turning these materials into paper are
$1⋅x for A4 and $3⋅y for A3.
There are different production procedures to produce each paper size. The production functions below give the number of reams of paper produced out of a given amount of materials.
f(x)g(y)=25x0.8=10y0.6(for A4)(for A3)
Build the (total) profit equation in terms of x and y. That is, find an equation Π(x,y) that gives the total profit (revenue minus cost) over both paper types.
Find the production quantities of both sizes of paper that maximizes profit.
If the branch stops producing A4, what is the optimal production for A3 to maximize profit?
Hint+
Instead of maximizing the total profit function, maximize the profit functions of each type of paper.
Answer+
The total profit is given by
Π(x,y)=(15x0.8−x)+(80y0.6−3y)
The optimal production: x=248,832 leading to 51840 reams of A4 and y=1,024 leading to 640 reams of A3
In this case, the optimal production is still 640 reams of A3
Full solution+
Let us first find the profit equations for each of the paper sizes separately and then we sum them up to get the total profit function.
Π4(x)Π3(y)=f(x)(6)−x(1)=15x0.8−x=g(y)(8)−y(3)=80y0.6−3y(profit for A4)(profit for A3)
and therefore, the total profit equation is given by
Π(x,y)=Π4(x)+Π3(y)=(15x0.8−x)+(80y0.6−3y)
Note that the production functions of the two paper types aren't really linked. It's as if one firm is doing all the A4, and a different firm is doing all the A3. So to maximize Π(x,y), we can just find the maximum value of Π4 and the maximum value of Π3 separately.
Note that x4,x3,x2≥0 as we cannot produce negative amount of papers. (Maybe that would mean turning papers into trees?) Note also:
(Also x=0 and y=0 are critical points, since the derivatives are undefined there, but we've already considered them when we thought about endpoints.)
Since Π4(125)>Π4(0) and Π3(210)>Π3(0), we see our maximum will occur when x=125 and y=210. Then the number of reams produced will be:
f(125)g(210)=25(125)4/5=51840=10(210)6/10=640
As we saw before, the two reams are optimized separately. So the optimal production of A3 isn't affected by how much A4 is produced. That is, the branch should stick with y=1,024 leading to 640 reams of A3.
Ayan and Pipe each have a lemonade boutique. Making each pitcher of lemonade costs $1. If Ayan wants to sell qA lemonades, and Pipe want to sell qP lemonades, then each pitcher of lemonade will be sold for this price:
p(qA,qP)=121−2(qA+qP)
Build the profit equation in terms of qA and qP for Ayan. Treating qP as a constant, find the value of qA that maximizes Ayan's profit. (Your answer will depend on qP.)
Build the profit equation in terms of qA and qP for Pipe. Treating qA as a constant, find the value of qP that maximizes Pipe's profit. (Your answer will depend on qA.)
Guess, using your intuition, how many pitchers are Ayan and Pipe are going to produce proportional to one another so that both of them maximize their respective profit functions.
Verify your answer for (c) mathematically.
Calculate the profit that each seller generates under these assumptions.
What would be their joint profit if they collaborate? Build a new profit equation where Ayan and Pipe are collaborating and find the optimal joint profit. Compare this to their individual profit when they are competing and decide whether it would be better for them to collaborate or compete.
Is it better for thirsty consumers when the two sellers collaborate, or when they compete?
Hint+
Profit is (revenue) minus (costs). If Ayan and Pipe work separately, then each seller only sees the cost and revenue from the lemonade that they themselves sold.
To find how much each seller will sell when they are working separately, find out which values of qA and qP end up with both individual profit functions being maximized.
To find out how much they'll sell when they're working together, use your assumption from part (c) to make the solving smoother.
Answer+
ΠA(qA)=−2qA2+120qA−2qAqP; maximum profit when qA=30−21qP
ΠP(qP)=−2qP2+120qP−2qPqA; maximum profit when qP=30−21qA
Their businesses are identical, so we predict they will sell the same amounts of lemonade.
If Ayan and Pipe sell 20 pitchers they will maximize their respective profit functions.
They would each make 800 dollars in profit.
Their optimal joint profit will be 1,800 dollars. But, they need to share this profit among the two of them. So if they collaborate, they will each earn 900 dollars. This is more than their individual optimal profit in the scenario where they are competing found in part (e) (we found this to be $800). So it is better for them to collaborate!
Collaborating sellers lead to higher prices and fewer goods, so it's better for consumers with the sellers compete
Full solution+
To find Ayan's profit equation, which we denote by ΠA, we just plug in the information we are given in the general profit equation (revenue minus cost).
Note that this is ΠA if we switch the places of qA and qP. So
Pipe would maximize their profit by selling 30−21qA pitchers of lemonade.
Ayan's and Pipe's cost and price for every pitcher of lemonade produced are the same. Their businesses are identical. So we predict that they will sell the same amount of lemonade to maximize their respective profits.
To find how much each seller will sell when they are working separately, find out which values of qA and qP end up with both individual profit functions being maximized.
Therefore we solve the system of equations we get from (a) and (b).
So, as predicted, both sellers sell the same number of pitchers.
We need to plug in qP=qA=20 in ΠA and ΠP:
ΠA(20)∣qP=20=−2(20)2+120(20)−2(20)(20)=800
And similarly, ΠP(20)∣qA=20=800. So, they would each make 800 dollars in profit.
The joint profit function is Π(qA,qP)=ΠA(qA)+ΠP(qP).
Note that here, Ayan and Pipe are helping each other to make the most profit, instead of competing. Using the same intuition as before, we can conclude that qA=qP in this case too. (So they share the workload fairly!)
So to make things easier let us assume qA=qP and denote this quantity by q. Then ΠA(q)=ΠP(q)=−4q2+120q. This means
This is a parabola pointing down, so its global min is at its sole critical point, q=15.
So q=qA=qP=15 maximizes the joint profit. Let us compute the corresponding joint profit
Πjoint(15)=−8(15)2+240(15)=1,800
So their optimal joint profit will be 1,800 dollars. But, they need to share this profit among the two of them. So if they collaborate, they will each earn 900 dollars. This is more than their individual optimal profit in the scenario where they are competing found in part (e) (we found this to be $800). So it is better for them to collaborate!
When the two sellers collaborate, they sell fewer lemonades (30 pitchers total instead of 40 total) and the lemonade costs more ($60 instead of $40). So it's better for consumers when the sellers compete.
