Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 6
Each set of homework questions is worth 100 marks
Due: at the beginning of the tutorial session Thursday/Friday, 10/11 March 2016
Name:
1. Consider the function
f (x, y) = x2 − xy 2 − 3x + y 4 + 5
Locate all relative maxima, relative minima, and saddle points, if any. Use Mathematica
to plot its graph.
Solution: We first find all critical points
fx (x, y) = 2x − y 2 − 3 = 0 ,
fy (x, y) = −2xy + 4y 3 = 0 .
From the first equation we find x in terms of y
x=
y2 3
+ ,
2
2
and substituting it to the second equation, we derive the following equation for y
3y 3 − 3y = 0 .
There are three solutions to this equation
y = 0 , y = −1 , y = 1 ,
and, therefore, three critical points
(x =
3
, y = 0) ,
2
(x = 2 , y = −1) ,
(x = 2 , y = 1) .
Computing the values of f at critical points, we get
11
3
,
f ( , 0) =
2
4
f (2, −1) = 2 ,
f (2, 1) = 2 .
To find out if they are maximum, minimum or saddle points we use the second derivative
test. To this end we compute
∂ 2f
∂ 2f
∂ 2f
2
(x,
y)
=
2
,
(x,
y)
=
−2x
+
12y
,
(x, y) = −2y ,
∂x2
∂y 2
∂x∂y
and
∂ 2f ∂ 2f
D(x, y) =
−
∂x2 ∂y 2
1
∂ 2f
∂x∂y
2
= 20y 2 − 4x ,
Computing D and
∂2f
∂x2
for the three critical points, we get
3
D( , 0) = −6 ,
2
∂ 2f 3
( , 0) = 2 ,
∂x2 2
and therefore (0 , 23 ) is a saddle point.
D(2, −1) = 12 ,
∂ 2f
(2, −1) = 2 ,
∂x2
and therefore (2, −1) is a relative minimum.
D(2, 1) = 12 ,
∂ 2f
(2, 1) = 2 ,
∂x2
and therefore (2, 1) is a relative minimum too.
The graph of the function is shown below
2. Consider the intersection of the surfaces
z=
p
a2 − x2 − y 2 , and
x2
y2
+
= 1,
b2
c2
a > b > c.
p
p
(a) What is the surface z = a2 − x2 − y 2 ? Sketch the surface z = a2 − x2 − y 2 and
its projection onto the xy plane for a = 3.
2
2
(b) What is the surface xb2 + yc2 = 1 ? Sketch the surface
onto the xy plane for b = 2, c = 1.
x2
b2
+
y2
c2
= 1 and its projection
(c) Use Lagrange multipliers to find the coordinates of the points on the intersection
which have the maximum z-coordinate and the minimum z-coordinate.
2
Solution :
(a) It is the upper semi-sphere of radius a. Its projection onto the xy plane for a = 3 is
a circle of radius 3.
(b) It is an elliptic cylinder. Its projection onto the xy plane for b = 2, c = 1 is an ellipse
with semi-axis 2 and 1.
(c) To find zmax nd zmin we use the Lagrange multiplier method, and get the equations
−
x
x
= 2Λ 2 ,
z
b
y
y
− = 2Λ 2 ,
z
c
z=
p
a2 − x 2 − y 2 ,
x2 y 2
+ 2 = 1.
b2
c
(1)
Since b 6= c these equations have four solutions
√
1),2) x = 0 , y = ±c , zmax = a2 − c2 .
√
3),4) x = ±b , y = 0 , zmin = a2 − b2 .
3. Let an open top box (a box without a lid) have volume 4m3 . Find the dimensions of the
box so that the area A of the box is minimised. What is the minimum area A?
Solution: The area of the box is
A(x, y, z) = xy + 2xz + 2yz .
(2)
where x, y, z are the length, width, and height of the box. The volume of a box is
V (x, y, z) = xyz = 4 ,
(3)
and it is the constraint.
The minimum of A is therefore given by
y + 2z = λyz ,
x + 2z = λxz ,
2x + 2y = λxy .
(4)
Multiplying the first equation by x and the second equation by y, and subtracting them
from each other, one gets
0 = 2zλ(x − y)
=⇒
y = x,
(5)
where we took into account that z = 0 or λ = 0 obviously cannot be solutions. Thus,
from the third equation, and then the second equation in (4) we get
4 = λx
=⇒
4z = x + 2z
=⇒
z=
x
.
2
(6)
Substituting these into the constraint equation one gets
x3 = 8
=⇒
x = 2, y = 2,z = 1.
(7)
Thus, the minimum area A is
A = 12m2 .
3
(8)
4. Show that a triangle with fixed area has minimum perimiter if it is equilateral.
Solution: Let the triangle has the sides a, b, c and the angle between the sides a and b is φ,
see the picture.
Then, the perimeter is
P (a, b, c, φ) = a + b + c ,
(9)
and we have the area constraint
1
A(a, b, c, φ) = ab sin φ − s = 0 ,
2
s is constant ,
(10)
and the constraint which relates a, b, c and φ is
Φ(a, b, c, φ) = a2 + b2 − 2ab cos φ − c2 = 0 .
(11)
We introduce two Lagrange multipliers λA and λΦ , and get the equations
1
1 = λA b sin φ + λΦ (2a − 2b cos φ) ,
2
1
1 = λA a sin φ + λΦ (2b − 2a cos φ) ,
2
1
1 = λΦ (−2c) =⇒ λΦ = − ,
2c
1
0 = λA ab cos φ + λΦ (2ab sin φ) =⇒
2
(12)
λA =
2
tan φ .
c
Substituting λP and λΦ into the first two equations one gets
a b 1
1=− +
c c cos φ
b a 1
1=− +
c c cos φ
=⇒
b = (c + a) cos φ ,
(13)
=⇒
a = (c + b) cos φ .
Subtracting the first equation from the second one, one gets
a − b = −(a − b) cos φ
=⇒
4
a = b,
cos φ =
a
,
a+c
(14)
because 0 < φ < π. Substituting the found values in Φ, one gets
a
2a2 (1 −
) − c2 = 0 =⇒ 2a2 = c(a + c) =⇒
a+c
c = a.
(15)
Thus, a = b = c, and the triangle is equilateral.
5. Consider the intersection of the surfaces
z = λ x + µ y + h , λ > 0 , µ > 0 , h > 0 , and
x2
y2
+ 2 = 1.
a2
b
(a) What is the surface z = λ x + µ y + h ? What is the surface
x2
a2
+
y2
b2
=1?
(b) Sketch the surfaces for a = b = λ = µ = h = 1.
(c) Use the Lagrange multiplier method to find the coordinates of the points on the
intersection which have the maximum z-coordinate and the minimum z-coordinate.
Solution :
(a) It is a plane. It is an elliptic cylinder.
(b)
(c) To find zmax and zmin we use the Lagrange multiplier method, and get the equations
x
λ = 2Λ 2 ,
a
y
µ = 2Λ 2 ,
b
x2 y 2
+ 2 = 1.
a2
b
(16)
Multiplying the first equation by x, the second equation by y and summing up the resulting equations, one gets
λx + µy = 2Λ .
(17)
Thus, one derives the equations for x and y
x
y
λ = (λx + µy) 2 , µ = (λx + µy) 2 ,
a
b
or by using the constraint
x2
a2
+
y2
b2
(18)
=1
y2
µ
= 2 xy ,
2
b
λa
5
x2
λ
= 2 xy .
2
a
µb
(19)
Summing up these equations one finds
λµa2 b2
,
λ 2 a2 + µ 2 b 2
Thus there are two solutions
xy =
x2 =
λ2 a4
,
λ 2 a2 + µ 2 b 2
λa2
,
xmin = − p
λ 2 a2 + µ 2 b 2
λa2
xmax = + p
,
λ 2 a2 + µ 2 b 2
y2 =
ymin = − p
ymax
µ 2 b4
.
λ 2 a2 + µ 2 b 2
(20)
µb2
,
λ2 a2 + µ2 b2
µb2
= +p
,
λ 2 a2 + µ 2 b 2
(21)
corresponding to the lowest and highest points on the intersection, respectively. The
z-coordinates of these points are
p
p
(22)
zmin = h − λ2 a2 + µ2 b2 , zmax = h + λ2 a2 + µ2 b2 .
Thus, the coordinates of the point on the intersection which has the maximum z-coordinate
are
p
λa2
µb2
xmax = + p
, ymax = + p
, zmax = h + λ2 a2 + µ2 b2 . (23)
λ2 a2 + µ2 b2
λ2 a2 + µ2 b2
while the coordinates of the point on the intersection which has the minimum z-coordinate
are
p
λa2
µb2
xmin = − p
, ymin = − p
, zmin = h − λ2 a2 + µ2 b2 . (24)
λ 2 a2 + µ 2 b 2
λ 2 a2 + µ 2 b 2
6. What is the volume of the largest n-dimensional box with edges parallel to the coordinate
axes that fits inside the n-dimensional ellipsoid
x2n
x21 x22
+
+
·
·
·
+
= 1.
a21 a22
a2n
(25)
Solution: The volume of a box with edges parallel to the coordinate axes that fits inside the
ellipsoid is
V (x1 , . . . , xn ) = 2n x1 · · · xn ,
(26)
where xi > 0 are coordinates of the vertex of the box in the first “octant”. The constraint
is
x2 x2
x2
(27)
g(x1 , . . . xn ) = 21 + 22 + · · · + 2n − 1 = 0 .
a1 a2
an
The maximum of V is therefore given by
x1 · · · xn
V
2xk
2n
=
= λ 2 , k = 1, . . . , n ,
(28)
xk
xk
ak
and therefore
x2k
ak
= V =⇒ 2λ = nV =⇒ xk = √ ,
(29)
2
ak
n
where we summed over k and used the constraint. Thus, the maximum volume V is
n
2
V = √
a1 · · · an .
(30)
n
2λ
6