Problem 1.

Find the critical points of the function

f(x, y) = 2x3 − 3x2y − 12x2 − 3y2

and determine their type i.e. local min/local max/saddle point. Are there any global min/max?

Solution: Partial derivatives

fx = 6x2 − 6xy − 24x, fy = −3x2 − 6y.

To find the critical points, we solve

fx = 0 =⇒ x2 − xy − 4x = 0 =⇒ x(x− y − 4) = 0 =⇒ x = 0 or x− y − 4 = 0

fy = 0 =⇒ x2 + 2y = 0.

When x = 0 we find y = 0 from the second equation. In the second case, we solve the system below bysubstitution

x− y − 4 = 0, x2 + 2y = 0 =⇒ x2 + 2x− 8 = 0

=⇒ x = 2 or x = −4 =⇒ y = −2 or y = −8.

The three critical points are(0, 0), (2,−2), (−4,−8).

To find the nature of the critical points, we apply the second derivative test. We have

A = fxx = 12x− 6y − 24, B = fxy = −6x, C = fyy = −6.

At the point (0, 0) we have

fxx = −24, fxy = 0, fyy = −6 =⇒ AC −B2 = (−24)(−6)− 0 > 0 =⇒ (0, 0)is local max .

Similarly, we find

(2,−2) is a saddle point

sinceAC −B2 = (12)(−6)− (−12)2 =< 0

and(−4,−8) is saddle

sinceAC −B2 = (−24)(−6)− (24)2 < 0.

The function has no global min since

limy→∞,x=0

f(x, y) = −∞

and similarly there is no global maximum since

limx→∞,y=0

f(x, y) =∞.

Problem 2.

Determine the global max and min of the function

f(x, y) = x2 − 2x+ 2y2 − 2y + 2xy

over the compact region−1 ≤ x ≤ 1, 0 ≤ y ≤ 2.

Solution: We look for the critical points in the interior:

∇f = (2x− 2 + 2y, 4y − 2 + 2x) = (0, 0) =⇒ 2x− 2 + 2y = 4y − 2 + 2x = 0 =⇒ y = 0, x = 1.

1

However, the point (1, 0) is not in the interior so we discard it for now.

We check the boundary. There are four lines to be considered:

• the line x = −1:

f(−1, y) = 3 + 2y2 − 4y.

The critical points of this function of y are found by setting the derivative to zero:

∂

∂y(3 + 2y2 − 4y) = 0 =⇒ 4y − 4 = 0 =⇒ y = 1 with f(−1, 1) = 1 .

• the line x = 1:

f(1, y) = 2y2 − 1.

Computing the derivative and setting it to 0 we find the critical point y = 0. The correspondingpoint (1, 0) is one of the corners, and we will consider it separately below.

• the line y = 0:

f(x, 0) = x2 − 2x.

Computing the derivative and setting it to 0 we find 2x − 2 = 0 =⇒ x = 1. This gives the corner(1, 0) as before.

• the line y = 2:

f(x, 2) = x2 + 2x+ 4

with critical point x = −1 which is again a corner.

Finally, we check the four corners

(−1, 0), (1, 0), (−1, 2), (1, 2).

The values of the function f are

f(−1, 0) = 3 , f(1, 0) = −1 , f(−1, 2) = 3 , f(1, 2) = 7 .

From the boxed values we select the lowest and the highest to find the global min and global max. Weconclude that

global minimum occurs at (1, 0)

global maximum occurs at (1, 2).

Problem 3.

Using Lagrange multipliers, optimize the function

f(x, y) = x2 + (y + 1)2

subject to the constraint

2x2 + (y − 1)2 ≤ 18.

Solution: We check for the critical points in the interior

fx = 2x, fy = 2(y + 1) =⇒ (0,−1) is a critical point .

The second derivative test

fxx = 2, fyy = 2, fxy = 0

shows this a local minimum with

f(0,−1) = 0 .

We check the boundary

g(x, y) = 2x2 + (y − 1)2 = 18

via Lagrange multipliers. We compute

∇f = (2x, 2(y + 1)) = λ∇g = λ(4x, 2(y − 1)).

Therefore

2x = 4xλ =⇒ x = 0 or λ =1

22(y + 1) = 2λ(y − 1).

In the first case x = 0 we get

g(0, y) = (y − 1)2 = 18 =⇒ y = 1 + 3√

2, 1− 3√

2

with values

f(0, 1 + 3√

2) = (2 + 3√

2)2 , f(0, 1− 3√

2) = (2− 3√

2)2 .

In the second case λ = 12 we obtain from the second equation

2(y + 1) = y − 1 =⇒ y = −3.

Now,

g(x, y) = 18 =⇒ x = ±1.

At (±1,−3), the function takes the value

f(±1,−3) = (±1)2 + (−3 + 1)2 = 5.

By comparing all boxed values, it is clear the (0,−1) is the minimum, while (0, 1 + 3√

2) is the maximum.

Problem 4.

Consider the function

w = ex2y

where

x = u√v, y =

1

uv2.

Using the chain rule, compute the derivatives

∂w

∂u,∂w

∂v.

Solution: We have

∂w

∂x= 2xy exp(x2y) = 2u

√v

1

uv2exp

(u2v · 1

uv2

)=

2

v3/2exp

(uv

)∂w

∂y= x2 exp(x2y) = u2v exp

(uv

)∂x

∂u=√v,

∂x

∂v=

u

2√v

∂y

∂u= − 1

u2v2,∂y

∂v= − 2

uv3.

Thus∂w

∂u=∂w

∂x· ∂x∂u

+∂w

∂y· ∂y∂u

=2

v3/2exp

(uv

)·√v − u2v exp

(uv

)· 1

u2v2=

=2

vexp

(uv

)− 1

vexp

(uv

)=

1

vexp

(uv

).

Similarly,∂w

∂u=∂w

∂x· ∂x∂u

+∂w

∂y· ∂y∂u

= − u

v2exp

(uv

).

Problem 5.

(i) Compute the degree n Taylor polynomial of the function

f(x, y) = ex2−y

around (0, 1).(ii) The second degree Taylor polynomial of a certain function f(x, y) around (0, 1) equals

1− 4x2 − 2(y − 1)2 + 3x(y − 1).

Can the point (0, 1) be a local minimum for f? How about a local maximum?

Solution:

(i) The Taylor series of the exponential function is

ez =∑n

zn

n!.

For z = x2 we obtain

ex2

=∑k

x2k

k!

and

e−y = e−1 · e−y+1 = e−1∑l

(−y + 1)l

l!.

Multiplying the degree n Taylor polynomial we need equals

e−1∑

2k+l≤n

x2k(−y + 1)l

k!l!.

(ii) From the Taylor polynomial we find

fx(0, 1) = fy(0, 1) = 0

so (0, 1) is a critical point. We can find the second derivatives

1

2fxx(0, 1) = −4,

1

2fyy(0, 1) = −2, fxy(0, 1) = 3.

By the second derivative test

AC −B2 = (−8)(−4)− 32 > 0, A = −8 < 0 =⇒ (0, 1) is a local maximum.

Problem 6.

Suppose that

z = e3x+2y, y = ln(3u− w), x = u+ 2v.

Calculate∂z

∂v,∂z

∂w.

Solution:

By the chain rule

∂z

∂v=∂z

∂x· ∂x∂v

= 3e3x+2y · 2 = 6e3xe2y = 6e3u+6ve2 ln(3u−w) = 6e3u+6v(3u− w)2.

Similarly,∂z

∂w=∂z

∂y· ∂y∂w

= 2e3x+2y · −1

3u− w= 2e3u+6ve2 ln(3u−w) −1

3u− w

= 2e3u+6v(3u− w)2 · −1

3u− w= −2e3u+6v(3u− w).

Problem 7.

Find the point on the plane

2x+ 3y + 4z = 29

that is closest to the origin.

Solution: We minimize the square of the distance to the origin

f(x, y, z) = x2 + y2 + z2

subject to the constraint

g(x, y, z) = 2x+ 3y + 4z = 29.

We use Lagrange multipliers

∇f = (2x, 2y, 2z),∇g = (2, 3, 4).

Then

∇f = λ∇g =⇒ 2x = 2λ, 2y = 3λ, 2z = 4λ =⇒ x = λ, y =3λ

2, z = 2λ.

Since

2x+ 3y + 4z = 2λ+9λ

2+ 8λ =

29λ

2= 29 =⇒ λ = 2 =⇒ x = 2, y = 3, z = 4.

The closest point is (2, 3, 4).

Problem 8.

Show that the function

f(x, y) =

{x3

x2+y2 if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)

admits all directional derivatives, it is continuous but it is not differentiable.

Solution: Let ~u = (u1, u2) be a unit direction. We calculate

f~u(0, 0) = limh→0

f((0, 0) + h~u)− f(0, 0)

h= lim

h→0

h3u31h2u21 + h2u22

· 1

h=

u31u21 + u22

.

This shows that the directional derivatives exist. Now the calculation also shows that

fx(0, 0) = 1, fy(0, 0) = 0

by specializing u to the unit vectors (1, 0) and (0, 1) respectively. If f is differentiable, then Df(0, 0) = (1, 0).However, in this case the directional derivative

f~u(0, 0) = Df(0, 0) · ~u = (1, 0) · ~u = u1

which does not agree with the answer above. Therefore, f is not differentiable.

To see that f is continuous, we observe that

|f(x, y)| = |x| · |x|2

|x|2 + |y|2≤ |x|

hence

lim(x,y)→(0,0)

f(x, y) = 0 = f(0, 0).

Problem 9.

Using Lagrange multipliers, find the minimal distance between two points on the ellipse x2 + 3y2 = 9 andthe circle x2 + y2 = 1.

Solution: We consider the points (a, b) and (c, d) on the ellipse and the circle such that

a2 + 3b2 = 9, c2 + d2 = 1.

The square of the distance between the two points equals

f(a, b, c, d) = (a− c)2 + (b− d)2.

Using Lagrange multipliers λ and µ we have to solve

2(a− c, b− d,−a+ c,−b+ d) = λ · 2(a, 3b, 0, 0) + µ · 2(0, 0, c, d).

(Note that the two gradients cannot be linearly dependent because of the zeros in their coordinates.) Thismeans

a− c = λa = −µc

b− d = 3λb = −µd.If λ = 0 then

a = c, b = d =⇒ a2 + 3b2 = 9, a2 + b2 = 1

which has no solutions.

When λ 6= 0, we obtain

a = tc, b =t

3d

where t = −µ/λ. The equations

a− c = −µc =⇒ tc− c = −µc =⇒ t− 1 = −µ or c = 0

b− d = −µd =⇒ t

3d− d = −µd =⇒ t

3− 1 = −µ or d = 0.

If c, d 6= 0, we obtain

t− 1 =t

3− 1 = −µ =⇒ t = 0.

This gives a = b = 0 which does not satisfy a2 + 3b2 = 9.

Otherwise, assume c = 0 which gives d = ±1 because c2 + d2 = 1. Since a− c = −µc we find a = 0 whichgives b2 = 3 so b = ±

√3. We find (0,±

√3) and (0,±1) as possible points with distance (

√3± 1)2.

When d = 0, we find c = ±1 and b = 0 by the same argument. Hence a = ±3. The points are (±3, 0) and(±1, 0) with distance either 4 or 10.

Out of all the values we found, the value (√

3− 1)2 is the smallest.

Problem 10.

Find the tangent planes of the following surfaces:

(i) x4yz + yz6 = 2 at (1, 1, 1);(ii) the graph of f(x, y) = x2y + 2xy3 at (1, 1, 3).

Solution:

(i) The gradient of the function f(x, y, z) = x4yz + yz6 equals

∇f = (4x3yz, x4z + z6, x4y + 6yz5)

which at (1, 1, 1) gives the normal vector ∇f(1, 1, 1) = (4, 2, 7). The tangent plane is

4(x− 1) + 2(y − 1) + 7(z − 1) = 0 .

(ii) We havefx = 2xy + 2y3 =⇒ fx(1, 1) = 4

fy = x2 + 6xy2 =⇒ fy(1, 1) = 7.

The tangent graph is

z − f(1, 1) = fx(1, 1)(x− 1) + fy(1, 1)(y − 1) =⇒ z − 3 = 4(x− 1) + 7(y − 1) .

Problem 11.

Consider p(x, y) a homogeneous degree 4 polynomial

p(x, y) = ax4 + bx3y + cx2y2 + dxy3 + ey4.

Let f : R2 → R be given by

f(x, y) =

{p(x,y)x2+y2 if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0).

Under what conditions on the coefficients a, b, c, d, e it is true that fxy(0, 0) and fyx(0, 0) exist and are equal.

Solution: By the quotient rule, we compute

fx =(x2 + y2) · px − 2x · p

(x2 + y2)2

for (x, y) 6= (0, 0). We havepx = 4ax3 + 3bx2y + 2cxy2 + dy3.

This yields the derivative

fx(0, h) =h2 · px(0, h)− 2 · 0 · p(0, h)

h4=px(0, h)

h2=dh3

h2= dh.

We will use this below.

The derivative at (0, 0) is calculated from the definition

fx(0, 0) = limh→0

f(h, 0)− f(0, 0)

h= lim

h→0

ah4

h2· 1

h= 0.

From here, we calculate fxy(0, 0) from the definition, as the derivative of fx in the y direction:

fxy(0, 0) = limh→0

fx(0, h)− fx(0, 0)

h= lim

h→0

dh− 0

h= d.

We similarly have fyx(0, 0) = b, by symmetry. So we need b = d.

Problem 12.

Consider the function f(x, y) = φ(

x−yx+y

). For what functions φ is is true that

fxx = fyy.

Solution: We calculate the derivatives via the chain rule. Write

z =x− yx+ y

.

Then∂z

∂x=

2y

(x+ y)2

∂z

∂y=

−2x

(x+ y)2.

Thus

fx = φ′(z) · ∂z∂x

= φ′(z) · 2y

(x+ y)2

fy = φ′(z) · ∂z∂y

= φ′(z) · −2x

(x+ y)2.

We use the product rule to find the second order derivatives. In the calculation of the second term below,we also used the chain rule once again. We have

fxx = φ′(z) · ∂∂x

(2y

(x+ y)2

)+ φ′′(z) · ∂z

∂x· 2y

(x+ y)2

= φ′(z) · −4y

(x+ y)3+ φ′′(z) · 2y

(x+ y)2· 2y

(x+ y)2.

SImilarly, we find

fyy = φ′(z) · 4x

(x+ y)3+ φ′′(z) · −2x

(x+ y)2· −2x

(x+ y)2.

Since fxx = fyy we obtain by subtracting the last two equations that

φ′(z) ·(

4x

(x+ y)3+

4y

(x+ y)3

)+ φ′′(z) ·

((−2x)(−2x)− (2y)(2y)

(x+ y)4

)= 0.

This expression simplifies as follows

φ′(z) · 4

(x+ y)2+ φ′′(z) · 4(x2 − y2)

(x+ y)4= 0

φ′(z) · 4

(x+ y)2+ φ′′(z) · 4(x− y)(x+ y)

(x+ y)4= 0.

Multiply by (x+ y)2 first, and then recall that x−yx+y = z. The above becomes

φ′(z) + φ′′(z) · z = 0.

Write ψ(z) = φ′(z). The above expression becomes

ψ(z) + ψ′(z) · z = 0 =⇒ ψ′(z)

ψ(z)= −1

z.

Integrating, we obtain

lnψ(z) = − ln(z) + C =⇒ ψ(z) = eC · 1

z=⇒ φ′(z) = eC · 1

z=⇒ φ(z) = eC ln(z) +B.

Writing A = eC , we see that the functions φ that satisfy the above condition are of the form

φ(z) = A ln(z) +B.