MY ACCOUNT

Calculus III
14.7: Maximum and Minimum Values
Study concepts, example questions & explanations

In almost all the problems in this sheet, we will use the Second
Derivative Test for functions of two variable to maximize and
minimize the given function.

NOTE 1: In case (c) the point (a, b) is called a saddle point of f and the
graph of f crosses its tangent plane at (a, b).

NOTE 2: If D = 0, the test gives no information (The Test Fails): could

have a local maximum or local minimum at (a, b), or could be a
saddle point of (a, b). In this case, we must find some other way to
determine the behavior of f at (a, b).

NOTE 3: We cannot apply this test if the partial derivatives of f are not
exist.
Question #1: Maximum and Minimum Values
Locate and classify the critical points of
𝒇(𝒙, 𝒚) = (𝒙 − 𝟓)𝟐 + (𝒚 + 𝟖)𝟐 .

Answer:

First compute the partial derivatives of f:

𝑓𝑥 = 2(𝑥 − 5), 𝑓𝑦 = 2(𝑦 + 8).

The critical points satisfy the equations 𝑓𝑥 = 0 and 𝑓𝑦 = 0

simultaneously. Hence

2(𝑥 − 5) = 0 ⟹ 𝑥=5

2(𝑦 + 8) = 0 ⟹ 𝑦 = −8

So, (5, −8) is the only critical point for f.

We now need to find D defined as

2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) .

To do so, we find the second partial derivatives

𝑓𝑥𝑥 = 2, 𝑓𝑦𝑦 = 2, 𝑓𝑥𝑦 = 0.

Thus,
2
𝐷(5, −8) = 𝑓𝑥𝑥 (5, −8)𝑓𝑦𝑦 (5, −8) − (𝑓𝑥𝑦 (5, −8)) = (2)(2) − (0)2 = 4.

Since 𝐷 > 0 and 𝑓𝑥𝑥 (2, −1) = 4 > 0, f has a local minimum at (5, −8)
and so 𝑓(5, −8) = 0 is a local minimum.
Question #2: Maximum and Minimum Values
Find the local maximum and minimum values and saddle point(s) of the
function
𝒇(𝒙, 𝒚) = 𝟐𝒙𝟐 + 𝟐𝒙𝒚 + 𝟐𝒚𝟐 − 𝟔𝒙.

Answer:

We first compute the partial derivatives of f:

𝑓𝑥 = 4𝑥 + 2𝑦 − 6, 𝑓𝑦 = 2𝑥 + 4𝑦.

The critical points satisfy the equations 𝑓𝑥 = 0 and 𝑓𝑦 = 0

simultaneously. Hence

4𝑥 + 2𝑦 − 6 = 0 ⟹ 4𝑥 + 2𝑦 = 6

2𝑥 + 4𝑦 = 0

The above system of equations has one solution at the point

(2, -1). So, (2, -1) is the only critical point for f.
We now need to find D defined as
2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) .

To do so, we find the second partial derivatives

𝑓𝑥𝑥 = 4, 𝑓𝑦𝑦 = 4, 𝑓𝑥𝑦 = 2.

Thus,
2
𝐷(2, −1) = 𝑓𝑥𝑥 (2, −1)𝑓𝑦𝑦 (2, −1) − (𝑓𝑥𝑦 (2, −1)) = (4)(4) − (2)2 = 12.

Since 𝐷 > 0 and 𝑓𝑥𝑥 (2, −1) = 4 > 0, f has a local minimum at (2, −1)
and so 𝑓(2, −1) = −6 is a local minimum.
Question #3: Maximum and Minimum Values
Determine the critical points and locate any local minima, maxima and
saddle points of the function
𝒇(𝒙, 𝒚) = 𝟐𝒙𝟐 − 𝟒𝒙𝒚 + 𝒚𝟒 + 𝟐

Answer:

We first compute the partial derivatives of f:

𝑓𝑥 = 4𝑥 − 4𝑦, 𝑓𝑦 = −4𝑥 + 4𝑦 3 .

The critical points satisfy the equations 𝑓𝑥 = 0 and 𝑓𝑦 = 0

simultaneously. Hence

4𝑥 − 4𝑦 = 0 ⟹ 𝑥 = 𝑦

−4𝑥 + 4𝑦 3 = 0

Substituting the first equation in the second equation gives

−4𝑦 + 4𝑦 3 = 0
4𝑦(𝑦 2 − 1) = 0
4𝑦(𝑦 − 1)(𝑦 + 1) = 0
𝑦 = 0, 𝑦 = 1, 𝑦 = −1.

We now use the equation x = y to find the critical points:

(0,0), (1,1), (−1, −1).

We now need to find D defined as

2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )

To do so, we find the second partial derivatives

𝑓𝑥𝑥 = 4, 𝑓𝑦𝑦 = 12𝑦 2 , 𝑓𝑥𝑦 = −4.

Thus,

(0,0) (1,1) (−1, −1)

𝑓𝑥𝑥 (𝑎, 𝑏) 4 4 4
𝑓𝑦𝑦 (𝑎, 𝑏) 0 0 0
𝑓𝑥𝑦 (𝑎, 𝑏) −4 −4 −4
𝐷 −16 32 32
Saddle point local minimum local minimum
Question #4: Maximum and Minimum Values
Find and classify the critical points of the function
𝟏 𝟑
+𝒙−𝒚𝟐 )
𝒇(𝒙, 𝒚) = 𝒆(−𝟑𝒙

Answer:

The partial derivatives of f are

1 3 1 3
+𝑥−𝑦 2 ) +𝑥−𝑦 2 )
𝑓𝑥 = (1 − 𝑥 2 )𝑒 (−3𝑥 , 𝑓𝑦 = −2𝑦𝑒 (−3𝑥 .

The critical points satisfy the equations 𝑓𝑥 = 0 and 𝑓𝑦 = 0

simultaneously. Hence
1 3
+𝑥−𝑦 2 )
(1 − 𝑥 2 )𝑒 (−3𝑥 =0 ⟹ (1 − 𝑥 2 ) = 0
1 3
+𝑥−𝑦 2 )
−2𝑦𝑒 (−3𝑥 =0 ⟹ 𝑦=0
The first equation gives 𝑥 = 1 𝑜𝑟 𝑥 = −1 and the second equation
gives 𝑦 = 0. Thus, the critical points are
(1,0), (−1,0).

We now need to find D defined as

2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) .

To do so, we find the second partial derivatives

1 3 1 3
+𝑥−𝑦 2 ) +𝑥−𝑦 2 )
𝑓𝑥𝑥 = (−2𝑥 + (1 − 𝑥 2 )2 )𝑒 (−3𝑥 , 𝑓𝑦𝑦 = (−2 + 4𝑦 2 )𝑒 (−3𝑥 ,

1 3
+𝑥−𝑦 2 )
𝑓𝑥𝑦 = −2𝑦(1 − 𝑥 2 )𝑒 (−3𝑥 .
Thus,

(1,0) (−1,0)
𝑓𝑥𝑥 (𝑎, 𝑏) −2𝑒 2/3 2𝑒 −2/3
𝑓𝑦𝑦 (𝑎, 𝑏) −2𝑒 2/3 −2𝑒 −2/3
𝑓𝑥𝑦 (𝑎, 𝑏) 0 0
𝐷 4𝑒 4/3 −4𝑒 −4/3
local maximum Saddle point
Question #5: Maximum and Minimum Values
Find and classify the critical points for
𝒇(𝒙, 𝒚) = 𝟑𝒙𝟐 𝒚 + 𝒚𝟑 − 𝟑𝒙𝟐 − 𝟑𝒚𝟐 + 𝟐

Answer:

We will first need to get all the first partial derivatives

𝑓𝑥 = 6𝑥𝑦 − 6𝑥, 𝑓𝑦 = 3𝑥 2 + 3𝑦 2 − 6𝑦.

The critical points satisfy the equations 𝑓𝑥 = 0 and 𝑓𝑦 = 0

simultaneously. Hence

6𝑥𝑦 − 6𝑥 = 0 ⟹ 6𝑥(𝑦 − 1) = 0
3𝑥 2 + 3𝑦 2 − 6𝑦 = 0

The first equation gives 𝑥 = 0 𝑜𝑟 𝑦 = 1. We plug these values in the

second equation to get
𝑥 = 0 ⟹ 3𝑦 2 − 6𝑦 = 3𝑦(𝑦 − 2) = 0 ⟹ 𝑦 = 0, 𝑦=2

𝑦=1 ⟹ 3𝑥 2 − 3 = 3(𝑥 − 1)(𝑥 + 1) = 0 ⟹ 𝑥 = 1, 𝑥 = −1

Thus, the critical points are

(0,0), (0,2), (1,1), (−1,1).

We now need to find D defined as

2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) .

To do so, we find the second partial derivatives

𝑓𝑥𝑥 = 6𝑦 − 6, 𝑓𝑦𝑦 = 6𝑦 − 6, 𝑓𝑥𝑦 = 6𝑥.

Thus,
(0,0) (0,2) (1,1) (−1,1)
𝑓𝑥𝑥 (𝑎, 𝑏) −6 6 0 0
𝑓𝑦𝑦 (𝑎, 𝑏) −6 6 0 0
𝑓𝑥𝑦 (𝑎, 𝑏) 0 0 6 −6
𝐷 36 36 −36 −36
local maximum local minimum Saddle point Saddle point
Question #6: Maximum and Minimum Values
Locate and classify the critical points of
𝒇(𝒙, 𝒚) = 𝟑𝒙𝒚 − 𝒙𝟑 − 𝒚𝟑 .

Answer:

First compute the partial derivatives of f:

𝑓𝑥 = 3𝑦 − 3𝑥 2 , 𝑓𝑦 = 3𝑥 − 3𝑦 2 .

The critical points satisfy the equations 𝑓𝑥 = 0 and 𝑓𝑦 = 0

simultaneously. Hence

3𝑦 − 3𝑥 2 = 0 ⟹ 𝑦 = 𝑥2

3𝑥 − 3𝑦 2 = 0 ⟹ 𝑥 = 𝑦2

The first equation gives 𝑦 = 𝑥 2 . We plug these values in the second

equation to get
2
𝑥 = 𝑦 2 = (𝑥2 ) = 𝑥4 ⟹ 𝑥4 − 𝑥 = 𝑥(𝑥3 − 1) = 0 ⟹ 𝑥 = 0 𝑜𝑟 1

Now, 𝑥 = 0 ⟹ 𝑦 = 02 = 0 and 𝑥 = 1 ⟹ 𝑦 = 12 = 1.
The critical points for f are
(0,0), (1,1).

We now need to find D defined as

2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) .

To do so, we find the second partial derivatives

𝑓𝑥𝑥 = −6𝑥, 𝑓𝑦𝑦 = −6𝑦, 𝑓𝑥𝑦 = 3.

Thus,
2
𝐷(0,0) = 𝑓𝑥𝑥 (0,0)𝑓𝑦𝑦 (0,0) − (𝑓𝑥𝑦 (0,0)) = (0)(0) − (3)2 = −9,
2
𝐷(1,1) = 𝑓𝑥𝑥 (1,1)𝑓𝑦𝑦 (1,1) − (𝑓𝑥𝑦 (1,1)) = (−6)(−6) − (3)2 = 27.

We can obtain that:

𝐷(0,0) < 0, and so (0,0) is a saddle point.
𝐷(1,1) > 0 with 𝑓𝑥𝑥 (2, −1) < 0 gives that f has a local maximum at
(1, 1).
Question #7: Applications on Extreme Values
Find the shortest distance from the point (1, 0, -2) to the plane
𝒙 + 𝟐𝒚 + 𝒛 = 𝟒.

Answer:

The distance from any point (𝑥, 𝑦, 𝑧) to the point (1, 0, -2) is

𝑑 = √(𝑥 − 1)2 + (𝑦 − 0)2 + (𝑧 + 2)2

but if (𝑥, 𝑦, 𝑧) lies on the plane 𝑥 + 2𝑦 + 𝑧 = 4, then

z = 4 − 𝑥 − 2𝑦 and so we have

𝑑 = √(𝑥 − 1)2 + (𝑦 − 0)2 + (4 − 𝑥 − 2𝑦 + 2)2 .

We can minimize d by minimizing the simpler expression

𝑑 2 = 𝑓(𝑥, 𝑦) = (𝑥 − 1)2 + 𝑦 2 + (6 − 𝑥 − 2𝑦)2 .

By solving the equations

𝑓𝑥 = 2(𝑥 − 1) − 2(6 − 𝑥 − 2𝑦) = 4𝑥 + 4𝑦 − 14 = 0,

𝑓𝑥 = 2𝑦 − 4(6 − 𝑥 − 2𝑦) = 4𝑥 + 10𝑦 − 24 = 0,

11 5
we find that the only critical point is ( , ).
6 3

Since 𝑓𝑥𝑥 = 4, 𝑓𝑥𝑦 = 4, 𝑓𝑦𝑦 = 10, we have

2
𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 ) = 24 > 0 and 𝑓𝑥𝑥 > 0,

11 5
so by the Second Derivatives Test f has a local minimum at ( , ).
6 3
Intuitively, we can see that this local minimum is actually an absolute
minimum because there must be a point on the given plane that is
5
closest to (1, 0, -2). At this point we get 𝑑 = √6.
6

So, the shortest distance from the point (1, 0, -2) to the plane
5
𝑥 + 2𝑦 + 𝑧 = 4 is √6.
6
Question #8: Finding Absolute Extrema
Find the absolute maximum and minimum values of the function
f (x, y) = x2 – 2xy + 2y

on the rectangle 𝑫 = {(𝒙, 𝒚)|𝟎 ≤ 𝒙 ≤ 𝟑, 𝟎 ≤ 𝒚 ≤ 𝟐}.

Answer:

Since f is a polynomial, it is continuous on the closed, bounded

rectangle D, so Theorem 8 tells us there is both an absolute
maximum and an absolute minimum.

We first find the critical points. These occur when

fx = 2x – 2y = 0, fy = –2x + 2 = 0,

so the only critical point is (1, 1), and the value of f there is f(1,1) = 1.

Next, we look at the values of f on the boundary of D, which consists

of the four line segments L1, L2, L3, L4 shown in the Figure below.

On L1 we have y = 0 and

f (x, 0) = x2 0  x  3.

This is an increasing function of x, so its minimum value is f(0,0) = 0

and its maximum value is f (3,0) = 9.

On L2 we have x = 3 and
 f(3, y) = 9 – 4y 0y2

This is a decreasing function of y, so its maximum value is f(3,0) = 9

and its minimum value is f(3,2) = 1.

On L3 we have y = 2 and

f(x, 2) = x2 – 4x + 4 0  x  3.

Simply by observing that f(x, 2) = (x – 2)2, we see that the minimum

value of this function is f(2,2) = 0 and the maximum value is f(0,2) =
4.

Finally, on L4 we have x = 0 and

f(0, y) = 2y 0y2

with maximum value f(0,2) = 4 and minimum value f(0,0) = 0.

Thus, on the boundary, the minimum value of f is 0 and the maximum

is 9.

Finally, we compare these values with the value f(1,1) = 1 at the

critical point and conclude that the absolute maximum value of f on D
is f(3, 0) = 9 and the absolute minimum value is f(0,0) = f(2, 2) = 0.

(𝑎, 𝑏) 𝑓(𝑎, 𝑏) Classification

Interior point(s) (1,1) 1
(0,0) 0 absolute minimum
(3,0) 9 absolute maximum
Boundary (3,2) 1
point(s)
(0,2) 4
(2,2) 0 absolute minimum

Question #9: : Finding Absolute Extrema

Find the absolute maximum and minimum values of
𝒇(𝒙, 𝒚) = 𝟐 + 𝟐𝒙 + 𝟐𝒚 − 𝒙𝟐 − 𝒚𝟐
on the triangular region in the first quadrant bounded by the lines x = 0,
y = 0, y = 9 − x.
Answer:

Since f is differentiable, the only places where f can assume these

values are points inside the triangle where 𝑓𝑥 = 𝑓𝑦 = 0 and points on
the boundary.

(a) Interior points. For these we have

𝑓𝑥 = 2 − 2𝑥 = 0, 𝑓𝑦 = 2 − 2𝑦 = 0,

yielding the single point (x, y) = (1, 1). The value of f there is f(1,1) =
4.

(b) Boundary points. We take the triangle one side at a time:

i) On the segment OA, y = 0. The function

𝑓(𝑥, 𝑦) = 𝑓(𝑥, 0) = 2 + 2𝑥 − 𝑥 2

may now be regarded as a function of x defined on the closed

interval [0, 9]. Its extreme values may occur at the endpoints

𝑥=0 𝑤ℎ𝑒𝑟𝑒 𝑓(0,0) = 2

𝑥=9 𝑤ℎ𝑒𝑟𝑒 𝑓(9,0) = −61

and at the interior points where 𝑓 ′ (𝑥, 0) = 2 − 2𝑥 = 0. The only

interior point where 𝑓 ′ (𝑥, 0) = 0 is x = 1, where f(1,0) = 3.
ii) On the segment OB, x = 0 and f(x,y) = f(0,y) = 2 + 2y - y2. We
know from the symmetry of f in x and y and from the analysis we just
carried out that the candidates on this segment are: 𝑓(0,0) = 2,
𝑓(0,9) = −61, 𝑓(0,1) = 3.

iii) We have already accounted for the values of f at the endpoints of

AB, so we need only look at the interior points of AB. With y = 9 − 𝑥,
we have

𝑓(𝑥, 9 − 𝑥) = 2 + 2𝑥 + 2(9 − 𝑥) − 𝑥 2 − (9 − 𝑥)2 = −61 + 18𝑥 − 2𝑥 2 .

Setting 𝑓 ′ (𝑥, 9 − 𝑥) = 18 − 4𝑥 = 0 gives 𝑥 = 9/2.

9
At this value of x, 𝑦 = 9 − 𝑥 = 9 − = 9/2 and
2

9 9
𝑓 ( , ) = −41/2.
2 2

We list all the candidates: 4, 2, − 61, 3, − 41/2. The absolute

maximum is 4, which f assumes at (1, 1). The absolute minimum is
− 61, which f assumes at (0, 9) and (9, 0).

(𝑎, 𝑏) 𝑓(𝑎, 𝑏) Classification

Interior point(s) (1,1) 4 absolute maximum
(1,0) 3
(0,1) 3
Boundary (0,0) 2
point(s)
(9,0) −61 absolute minimum
(0,9) −61 absolute minimum
9 9 41
( , ) −
2 2 2
Question #10: Finding Absolute Extrema
Find the absolute minimum and absolute maximum of
𝒇(𝒙, 𝒚) = 𝟐𝒙𝟐 − 𝒚𝟐 + 𝟔𝒚
on the disk 𝒙𝟐 + 𝒚𝟐 ≤ 𝟏𝟔.

Answer:

Since f is differentiable, the only places where f can assume these

values are points inside the disk and points on the boundary of the
disk.

Let’s first find the critical points of the function that lies inside the
disk. To do so, we solve

𝑓𝑥 = 4𝑥 = 0, 𝑓𝑦 = −2𝑦 + 6 = 0,

yielding the single point (0, 3). The value of f there is f(0, 3) = 9.

Now we need to look at the boundary. On the boundary we have

𝒙𝟐 + 𝒚𝟐 = 𝟏𝟔 and so 𝒙𝟐 = 𝟏𝟔 − 𝒚𝟐 . If we plug this in the rule of 𝒇(𝒙, 𝒚)
we get that

𝑓(𝑥, 𝑦) = 2𝑥 2 − 𝑦 2 + 6𝑦 = 2(16 − 𝑦2 ) − 𝑦 2 + 6𝑦

This is a function of y,

𝑔(y) = 32 − 3𝑦 2 + 6𝑦

We will need to find the absolute extrema of this function on the

range −4 ≤ 𝑦 ≤ 4 (this is the range of y on the disk).

Note that
𝑔′(𝑦) = −6y + 6 = 0 ⇛ 𝑦 = 1.
The value of this function at the critical point and the endpoints are,

𝑔(−4) = −40, 𝑔(1) = 35, 𝑔(−4) = 8.

To find the values of x that correspond to these values of y, we use

that 𝒙𝟐 = 𝟏𝟔 − 𝒚𝟐 . Thus,

𝑦 = −4 ⇛ 𝑥 = 0, 𝑦 = 1 ⇛ 𝑥 = ±√15, 𝑦 = 4 ⇛ 𝑥 = 0.

We then can find that

𝑓(0, −4) = −40, 𝑓(0,4) = 8, 𝑓(√15, 1) = 35, 𝑓(−√15, 1) = 35.

So, comparing these values to the value of the function at the critical
point, f(0, 3) = 9, we can see that the absolute minimum occurs
at (0,−4) while the absolute maximum occurs twice at (√15, 1) and
(−√15, 1).

(𝑎, 𝑏) 𝑓(𝑎, 𝑏) Classification

Interior point(s) (0,3) 9
Boundary (0, −4) −40 absolute minimum
point(s) (0,4) 8
(√15, 1) 35 absolute maximum

(−√15, 1) 35 absolute maximum