Calculus Volume 3 · Chapter 4 · Section 4.7

Maxima/Minima Problems

Where a surface tops out, bottoms out, or does both at once — and how to find the highest and lowest points of a region.

Idea 1 · flat in every directionExtrema sit at critical points.Both partials zero, or one of them undefined. Idea 2 · bowl, cap or saddleThe discriminant D decides.D > 0: a bowl or a cap. D < 0: a saddle. Idea 3 · check the edgeOn a closed region, the boundary counts.The largest candidate wins, wherever it sits.

Colour key minimum maximum saddle boundary key idea · 3D figures turn when you drag them · page links open the textbook

L1The top of a hill
L1Flat in every direction
L1Bowl, cap, saddle
L1One picture
L2Finding critical points
L2The second derivative test
L2Absolute extrema on a closed region
L2Examples 4.38–4.41
L2Lab 1 · read D anywhere
L2Lab 2 · walk the boundary
L2Lab 3 · golf-ball profit
L2Common mistakes
L2Practice and answers
L3Why D works
L3When D = 0
L3Why closed and bounded
L3Where it is used next

Level 1 · see itWhat a maximum, a minimum and a saddle look likebefore any test

The top of a hill

At the top of $z=\sqrt{16-x^2-y^2}$ (Figure 4.47), walk east or north: the ground is level either way.

Both slices through the top have slope 0, so $f_x=f_y=0$ there, and the tangent plane is horizontal.

A smooth top or bottom is a place where the surface is flat in every direction.

The lowest value, 0, is on the rim $x^2+y^2=16$ — the edge of the domain, where the surface is not flat at all.

The two slices through the top, and the flat tangent plane. Drag to turn.

Flat in every direction

Critical point · p387

$(x_0,y_0)$ is a critical point of $f$ if $f_x(x_0,y_0)=f_y(x_0,y_0)=0$ , or if $f_x$ or $f_y$ does not exist there.

Fermat's theorem (Theorem 4.16): a local maximum or minimum, where the partials exist, is at a critical point. Critical points are the only candidates — but not every candidate is an extremum.

z=x^2+y^2

f_x=f_y=0

at the bottom.

z=\sqrt{x^2+y^2}

: no partials at the tip, still a minimum.

Bowl, cap, saddle

Minimum:

x^2+y^2

D=4

Maximum:

-x^2-y^2

D=4

Saddle:

x^2-y^2

D=-4

All three have $f_x=f_y=0$ at the origin. The saddle (Figure 4.48) rises along the $x$ -slice $z=x^2$ and falls along the $y$ -slice $z=-y^2$ : a minimum one way, a maximum the other.

Contours tell them apart: closed rings round a top or bottom, crossing lines at a saddle.

One picture

Example 4.39b,

g=\tfrac13x^3+y^2+2xy-6x-3y+4

: a saddle at

(-1,\tfrac52)

D=-8

, and a minimum at

(3,-\tfrac32)

D=8

Find where the surface is flat; the discriminant says which kind of flat.

Level 2 · compute itCritical points, the test, the boundaryand three labs

Finding critical points

Recipe · p387

f_x(x,y)=0,\qquad f_y(x,y)=0

Solve the two equations together; then list any points where a partial derivative fails to exist.

Example 4.38b: $g=x^2+2xy-4y^2+4x-6y+4$

\begin{aligned} g_x&=2x+2y+4=0\\ g_y&=2x-8y-6=0 \end{aligned}\quad\Longrightarrow\quad (-1,-1),\ \ g(-1,-1)=5

Example 4.38a: $f=\sqrt{4y^2-9x^2+24y+36x+36}$ has $(2,-3)$ , and also every point of the hyperbola $\frac{(x-2)^2}{4}-\frac{(y+3)^2}{9}=1$ , where the partials do not exist.

The second derivative test

Theorem 4.17 · p390

D=f_{xx}(x_0,y_0)\,f_{yy}(x_0,y_0)-\big(f_{xy}(x_0,y_0)\big)^2

at a critical point where $f_x=f_y=0$ .

D	$f_{xx}$	at $(x_0,y_0)$
$D>0$	$>0$	local minimum
$D>0$	$<0$	local maximum
$D<0$	any	saddle point
$D=0$	any	no conclusion

D > 0 means the surface curves the same way in every direction; the sign of $f_{xx}$ says up or down. Steps (p391): critical points, then $D$ at each, then the table.

Absolute extrema on a closed region

Theorems 4.18–4.19 · p394

A continuous $f$ on a closed, bounded set $D$ has an absolute maximum and minimum, each at a critical point inside $D$ or on the boundary.

Critical points inside $D$ ; evaluate $f$ .
Each boundary piece as a one-variable function; its critical points and endpoints.
Largest value: absolute max. Smallest: absolute min.

Example 4.40a: every candidate and its value. 36 at

(0,2)

beats all; 17 at

(3,1)

is lowest.

On a closed region the winner is often on the edge, where $f_x$ and $f_y$ need not be zero.

Examples 4.38–4.41

Example	given	result
4.38a · p387	$\sqrt{4y^2-9x^2+24y+36x+36}$	$(2,-3)$ ; partials undefined on $\frac{(x-2)^2}{4}-\frac{(y+3)^2}{9}=1$
4.38b · p388	$x^2+2xy-4y^2+4x-6y+4$	$(-1,-1)$ , value 5
4.39a · p391	$4x^2+9y^2+8x-36y+24$	$D=144$ : minimum $-16$ at $(-1,2)$
4.39b · p393	$\tfrac13x^3+y^2+2xy-6x-3y+4$	$D=4x_0-4$ : saddle $(-1,\tfrac52)$ , minimum $-\tfrac{29}{4}$ at $(3,-\tfrac32)$
4.40a · p395	$x^2-2xy+4y^2-4x-2y+24$ on $[0,4]\times[0,2]$	max 36 at $(0,2)$ ; min 17 at $(3,1)$
4.40b · p397	$x^2+y^2+4x-6y$ on $x^2+y^2\le16$	min $-13$ at $(-2,3)$ ; max $16+8\sqrt{13}\approx44.84$ on the circle
4.41 · p399	$48x+96y-x^2-2xy-9y^2$ on $[0,50]\times[0,25]$	max 648 at $(21,3)$ : $648,000

Figure 4.55 (p398) labels its points $(\tfrac{12}{5},-\tfrac{16}{5},\tfrac{224}{5})$ and $(-\tfrac{12}{5},\tfrac{16}{5},-\tfrac{64}{5})$ ; the solution's values, 44.84 on the circle and $-13$ at $(-2,3)$ , are the right ones.

Lab 1 · read D anywhere

surface

Drag the dot on the contour map, or jump to a critical point. Away from a critical point the tangent plane tilts; at one it lies flat. Try x³ − 3xy²: $D=0$ at the origin and the test says nothing, yet it is a saddle with three valleys.

Lab 2 · walk the boundary

problem

walk the edge, s0

The right-hand graph is $f$ along the whole boundary, one piece after another. Its highs and lows, the corners, and the inside critical point are the only places the answer can be.

Lab 3 · golf-ball profit

Example 4.41: $x$ thousand balls, $y$ hours of advertising, profit $f=48x+96y-x^2-2xy-9y^2$ thousand dollars, on $0\le x\le 50$ , $0\le y\le 25$ .

Surface drawn for

0\le y\le 12

; past that the profit drops steeply.

Drag the plan. At the best plan both slices are flat: one more ball or one more hour gains nothing. The edges top out at 576 (no advertising) and 256 (no balls), below 648.

Common mistakes

Mistake	Result	Fix
$D<0$ read as a maximum	a saddle called a top	$D<0$ : saddle; only $D>0$ gives an extremum
Deciding max/min from $D$ alone	Checkpoint 4.35: $D=68$ is a maximum, not a minimum	look at $f_{xx}$ when $D>0$
$D=0$ read as "no extremum"	Exercise 325: $9-x^4y^4$ has a maximum at the origin	$D=0$ : the test is silent; look at $f$
Skipping points where a partial fails to exist	the tip of $\sqrt{x^2+y^2}+1$ missed (Exercise 317)	list them as critical points
Only interior critical points on a closed region	Example 4.40a: 17 found, the maximum 36 on the edge missed	every edge, and the corners
Reporting the parameter $t$ , not the point	" $t=\tfrac54$ " for the point $(4,\tfrac54)$	substitute back into $x(t),y(t)$

Practice and answers

Exercises from p401; the book's key: p864, p870–871.

Checkpoints 4.34–4.36

#	task	answer
4.34	critical point of $x^3+2xy-2x-4y$	$(2,-5)$
4.35	local extrema of $x^3+2xy-6x-4y^2$	saddle $(\tfrac43,\tfrac13)$ , $D=-68$ ; local max $(-\tfrac32,-\tfrac38)$ , $D=68$ , value $\tfrac{99}{16}$
4.36	absolute extrema of $4x^2-2xy+6y^2-8x+2y+3$ on $[0,2]\times[-1,3]$	min $f(1,0)=-1$ ; max $f(0,3)=63$

Exercises

#	function	answer
311	$(3x-2)^2+(y-4)^2$	$(\tfrac23,4)$
313	$15x^3-3xy+15y^3$	$(0,0)$ , $(\tfrac1{15},\tfrac1{15})$
315	$-x^2-5y^2+8x-10y-13$	maximum at $(4,-1,8)$
317	$\sqrt{x^2+y^2}+1$	minimum at $(0,0,1)$
319	$x^2y^2$	test fails; absolute minimum 0 on the axes
321	$2xy+3x+4y$	saddle $f(-2,-\tfrac32)=-6$
323	$x^2+4xy+y^2$	saddle $(0,0,0)$
325	$9-x^4y^4$	$D=0$ ; a maximum, $f(0,0)=9$
327	$3x^2-2xy+y^2-8y$	minimum at $(2,6)$
329	$y^2+xy+3y+2x+3$	saddle $(1,-2)$
331	$x^2+2y^2-x^2y$	saddles $(\pm2,1)$ ; minimum $(0,0)$
333	$e^{-(x^2+y^2+2x)}$	maximum at $(-1,0)$
335	$x^2+10xy+y^2$	saddle $(0,0)$
337	$120x+120y-xy-x^2-y^2$	maximum at $(40,40)$
339	$x^2+x-3xy+y^3-5$	saddle $(\tfrac14,\tfrac12)$ ; minimum $(1,1)$
341	$ye^x-e^y$	saddle $(0,0)$
343	$\sin x\sin y$ on $(0,2\pi)^2$	saddle $(\pi,\pi)$ ; max $(\tfrac\pi2,\tfrac\pi2),(\tfrac{3\pi}2,\tfrac{3\pi}2)$ ; min $(\tfrac\pi2,\tfrac{3\pi}2),(\tfrac{3\pi}2,\tfrac\pi2)$
345	$x^2+y^2-2y+1$ on $x^2+y^2\le4$	min $(0,1,0)$ ; max $(0,-2,9)$
347	$\frac{-2y}{x^2+y^2+1}$ on $x^2+y^2\le4$	min $(0,1,-1)$ ; max $(0,-1,1)$
349	points of $x^2-yz=5$ nearest the origin	$(\pm\sqrt5,0,0)$
351	largest package, length + girth 108 in	18 by 36 by 18 in
353	$z=x^2+y^2+10$ nearest $x+2y-z=0$	$(\tfrac12,1,\tfrac{45}4)$ ; on the plane $(\tfrac{47}{24},\tfrac{47}{12},\tfrac{235}{24})$
355	shoe revenue $R(x,y)$	$x=3$ , $y=6$
357	soda can, height + circumference 120 cm	$V=\tfrac{64{,}000}{\pi}\approx20{,}372\ \text{cm}^3$

Level 3 · why it worksWhere the test comes fromshort arguments

Why D works

Near a critical point, with $A=f_{xx}$ , $B=f_{xy}$ , $C=f_{yy}$ there, $f$ is its quadratic part:

f(x_0+h,y_0+k)\approx f(x_0,y_0)+\tfrac12\left(Ah^2+2Bhk+Ck^2\right)

Complete the square (for $A\ne0$ ):

Ah^2+2Bhk+Ck^2=A\left[\Big(h+\frac BA k\Big)^2+\frac{AC-B^2}{A^2}\,k^2\right]

If $D=AC-B^2>0$ , the bracket is positive for every direction, so $f$ rises all round ( $A>0$ ) or falls all round ( $A<0$ ). If $D<0$ , the bracket is positive along $k=0$ and negative along $h=-\frac BAk$ : a saddle. D is the discriminant of the quadratic: it asks whether the surface bends the same way in every direction.

When D = 0

$f$	$D(0,0)$	at the origin
$x^4+y^4$	0	minimum
$-x^4-y^4$	0	maximum
$x^4-y^4$	0	saddle

All three have zero second derivatives at the origin, so the quadratic part is 0 and says nothing. The test only reads the quadratic part; when it vanishes, higher terms decide.

Why closed and bounded

Not bounded

$x^2+y^2$ on the whole plane has no maximum: it grows without end.

Not closed

$x^2+y^2$ on the open disc $x^2+y^2<1$ approaches 1 but never reaches it: no maximum.

The extreme value theorem (Theorem 4.18) needs both. Then an absolute extremum inside $D$ is also a local one, so it is a critical point (Fermat); otherwise it is on the boundary. That is Theorem 4.19.

Where it is used next

Where	What changes
§4.8 Lagrange multipliers	a curved boundary $g(x,y)=c$ without a parametrization: $\nabla f=\lambda\nabla g$
Least squares	minimizing a sum of squares by setting partials to 0
Machine learning	gradient descent walks downhill to a critical point; saddles slow it down

Every method is the same two steps: find where the surface is flat, then check the edges.

Objective	You can now
4.7.1 critical points	solve $f_x=0,\ f_y=0$ ; find where a partial is undefined
4.7.2 second derivative test	$D=f_{xx}f_{yy}-f_{xy}^2$ : min, max, saddle, or no conclusion
4.7.3 absolute extrema	check interior critical points, each boundary piece and its corners

The top of a hill

Flat in every direction

Bowl, cap, saddle

One picture

Finding critical points

The second derivative test

Absolute extrema on a closed region

Examples 4.38–4.41

Lab 1 · read D anywhere

Lab 2 · walk the boundary

Lab 3 · golf-ball profit

Common mistakes

Practice and answers

Why D works

When D = 0

Why closed and bounded

Where it is used next

Maxima/Minima Problems

Candidates for a max or min: both partials are 0, or one does not exist

Where the partials' denominator is 0, every point is critical: a whole hyperbola

For a polynomial, set both partials to 0 and solve the system: (−1,−1)(-1,-1)(−1,−1)

A local max beats every nearby value; a global max beats every value in the domain

At a smooth local max or min, both partials are 0: the tangent plane is level

Fermat in two variables is Fermat in one variable, used twice

A critical point need not be an extremum: a saddle rises one way and falls another

The discriminant DDD sorts a critical point: minimum, maximum or saddle

D>0D>0D>0 means the surface bends the same way in every direction

When D=0D=0D=0 the test is silent: the same zero hides a min, a max or a saddle

One critical point, D=144>0D=144>0D=144>0 and fxx>0f_{xx}>0fxx​>0: a minimum of −16-16−16

DDD can change from point to point: one saddle, one minimum

One surface, two level spots: the sign of DDD tells the saddle from the minimum

A saddle at (43,13)(\tfrac43,\tfrac13)(34​,31​) and a local maximum at (−32,−38)(-\tfrac32,-\tfrac38)(−23​,−83​)

On a closed, bounded region a continuous function always reaches a max and a min

The absolute extremes are at interior critical points or on the boundary: compare them all

Theorem 4.19 is EVT plus Fermat; drop closed or bounded and the max can vanish

On the rectangle the max 36 is at a corner, the min 17 inside

On a disc, parametrize the circle: max ≈ 44.84 on the edge, min −13 inside

Min −1-1−1 inside at (1,0)(1,0)(1,0), max 63 at the corner (0,3)(0,3)(0,3)

Maximum profit $648,000: 21,000 balls and 3 hours of advertising a month

Critical points find the candidates, DDD sorts them, the boundary finishes the job

For a polynomial, set both partials to 0 and solve the system: $(-1,-1)$

The discriminant $D$ sorts a critical point: minimum, maximum or saddle

$D>0$ means the surface bends the same way in every direction

When $D=0$ the test is silent: the same zero hides a min, a max or a saddle

One critical point, $D=144>0$ and $f_{xx}>0$ : a minimum of $-16$

$D$ can change from point to point: one saddle, one minimum

One surface, two level spots: the sign of $D$ tells the saddle from the minimum

A saddle at $(\tfrac43,\tfrac13)$ and a local maximum at $(-\tfrac32,-\tfrac38)$

Min $-1$ inside at $(1,0)$ , max 63 at the corner $(0,3)$

Critical points find the candidates, $D$ sorts them, the boundary finishes the job