14b. Maximum and Minimum Values — Practice

Drill sheet for 14. Maximum and Minimum Values. Book-style examples above each worksheet; regenerate for new numbers. Proofs use guided choice.

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1. Absolute extrema on natural domains

Book. (a) $y=-x^{2}$: abs max $0$ at $x=0$; no abs min. (b) $y=(x-3)^{2}$: abs min $0$ at $x=3$; no abs max. (c) $y=\sqrt{25-4x^{2}}$: abs max $5$ at $0$, abs min $0$ at $\pm 5/2$. (d) $y=\sqrt{x-4}$: abs min $0$ at $x=4$; no abs max.

Absolute max/min on a domain

Parabolas, semicircles, square roots — read range and endpoints

(1)Locate absolute max/min on the natural domain: (a) $y=-x^{2}$ (b) $y=(x-3)^{2}$ (c) $y=\sqrt{25-4x^{2}}$ (d) $y=\sqrt{x-4}$.

(a) Parabola down: abs max $0$ at $x=0$; no abs min (range $(-\infty,0]$).

(b) Parabola up: abs min $0$ at $x=3$; no abs max.

(c) Upper semicircle/ellipse: abs max $5$ at $x=0$; abs min $0$ at $x=\pm 5/2$.

(d) Abs min $0$ at $x=4$; no abs max.

(a) max $0$ at $0$. (b) min $0$ at $3$. (c) max $5$ at $0$, min $0$ at $\pm 5/2$. (d) min $0$ at $4$.

(2)Locate absolute extrema of $y=-3x^{2}$ on $\mathbb{R}$.

$y\le 0$ with equality only at $x=0$.

Opens downward — abs max, no abs min.

Abs max $0$ at $x=0$; no absolute minimum.

(3)Locate absolute extrema of $y=\sqrt{x-1}$ on its domain.

Domain $[1,\infty)$. Value $0$ only at $x=1$; grows without bound.

Abs min $0$ at $x=1$; no absolute maximum.

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2. Cubic — critical numbers, second-derivative test, mono

Book. $f(x)=\dfrac{1}{3}x^{3}+\dfrac{1}{2}x^{2}-6x+8$. Critical $-3,\,2$; rel max at $-3$ ($f=\tfrac{43}{2}$); rel min at $2$ ($f=\tfrac{2}{3}$); increasing on $(-\infty,-3)$ and $(2,\infty)$, decreasing on $(-3,2)$. Neither absolute max nor min on $\mathbb{R}$.

Fig. 14-5 — cubic with a relative max and a relative min (no absolute extrema on ℝ)

Cubic extrema analysis

Factor f′, second-derivative test, sign chart for mono

(1)For a cubic with $f'(x)=(x-(-4))(x-(2))$, find critical numbers, use $f''(x)=2x-(-2)$ to classify, and state mono intervals.

Critical numbers: $-4$ and $2$.

$f''(-4)=-6$ <0 ⇒ rel max; $f''(2)=6$ >0 ⇒ rel min.

Sign of $f'$: increasing on $(-\infty,-4)$ and $(2,\infty)$; decreasing on $(-4,2)$.

Crit $-4,2$; classify via $f''$; mono intervals as in steps.

(2)For a cubic with $f'(x)=(x-(-2))(x-(2))$, find critical numbers, use $f''(x)=2x-(0)$ to classify, and state mono intervals.

Critical numbers: $-2$ and $2$.

$f''(-2)=-4$ <0 ⇒ rel max; $f''(2)=4$ >0 ⇒ rel min.

Sign of $f'$: increasing on $(-\infty,-2)$ and $(2,\infty)$; decreasing on $(-2,2)$.

Crit $-2,2$; classify via $f''$; mono intervals as in steps.

(3)For a cubic with $f'(x)=(x-(-4))(x-(1))$, find critical numbers, use $f''(x)=2x-(-3)$ to classify, and state mono intervals.

Critical numbers: $-4$ and $1$.

$f''(-4)=-5$ <0 ⇒ rel max; $f''(1)=5$ >0 ⇒ rel min.

Sign of $f'$: increasing on $(-\infty,-4)$ and $(1,\infty)$; decreasing on $(-4,1)$.

Crit $-4,1$; classify via $f''$; mono intervals as in steps.

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3. Quartic — three critical numbers

Book. $f(x)=x^{4}+2x^{3}-3x^{2}-4x+4$. Factor $f'(x)=2(x-1)(2x+1)(x+2)$; critical $1,-\tfrac{1}{2},-2$. Second-derivative test: mins at $1$ and $-2$; max at $-\tfrac{1}{2}$.

Fig. 14-6 — three critical numbers; two mins and one max

Quartic / cubic-derivative sketch

List critical numbers, classify, state mono intervals

(1)For $f(x)=x^{4}+2x^{3}-3x^{2}-4x+4$ with $f'(x)=2(x-1)(2x+1)(x+2)$: list critical numbers, classify via $f''$, state mono intervals.

Critical: $1$, $-1/2$, $-2$.

$f''(1)>0$ min; $f''(-1/2)<0$ max; $f''(-2)>0$ min.

Inc on $(-2,-1/2)$ and $(1,\infty)$; dec on $(-\infty,-2)$ and $(-1/2,1)$.

Crit $-2,-1/2,1$; mins at $-2,1$; max at $-1/2$; mono as above.

(2)Suppose $f'(x)=(x-(-2))(x-(0))(x-(2))$ with three distinct real zeros. List critical numbers and the mono pattern from the sign chart of $f'$.

Critical numbers: $-2$, $0$, $2$ (assume ordered $-2<0<2$).

Cubic factor with positive leading coeff: $f'$ is $-$ left of leftmost root, then alternates.

Inc where $f'>0$, dec where $f'<0$.

Critical $-2,0,2$; sign-chart $f'$ for mono intervals.

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4–5, 7. Special critical numbers

Book 4. $f(x)=1/(x-2)$: $f'<0$ wherever defined; no critical numbers; decreasing on each side of $x=2$.

Book 5. $f(x)=2+x^{2/3}$: cusp critical at $0$; $\{-,+\}$ ⇒ absolute min.

Book 7. $f(x)=(x-2)^{2/3}$: $f'$ undefined at $2$; first-derivative test ⇒ relative min.

Fig. 14-7 — no relative extrema; vertical asymptote

Fig. 14-8 — absolute minimum where f′ is undefined

Cusps, poles, first-derivative test

Critical when f′=0 or undefined — then read signs

(1)Determine relative extrema of $f(x)=(x-2)^{2/3}$ (first-derivative test).

$f'=\dfrac{2}{3}(x-2)^{-1/3}$; critical at $2$ ($f'$ undefined).

For $x<2$, $f'<0$; for $x>2$, $f'>0$ ⇒ $\{-,+\}$ ⇒ relative min at $2$.

Relative minimum at $x=2$.

(2)Locate relative extrema of $f(x)=2+x^{2/3}$ and mono intervals.

$f'=\dfrac{2}{3}x^{-1/3}$; critical at $0$ ($f'$ undefined).

$f'<0$ for $x<0$, $f'>0$ for $x>0$ ⇒ $\{-,+\}$ ⇒ abs/rel min at $0$.

Absolute (and relative) minimum at $x=0$; dec on $(-\infty,0)$, inc on $(0,\infty)$.

(3)Examine $f(x)=\dfrac{1}{x-2}$ for relative extrema and mono intervals.

$f'=-(x-2)^{-2}<0$ wherever defined; never $0$.

$x=2$ is not in the domain ⇒ no critical numbers ⇒ no relative extrema.

Decreasing on $(-\infty,2)$ and on $(2,\infty)$ separately.

No relative extrema; decreasing on each side of the asymptote $x=2$.

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6. Second-derivative test apps

Book. (a) $f(x)=x(12-2x)^{2}$: rel max at $2$, rel min at $6$. (b) $f(x)=x^{2}+250/x$: rel min at $x=5$.

Second-derivative applications

Expand, find critical numbers, sign of f″

(1)Use the second-derivative test on $f(x)=x^{2}+\dfrac{250}{x}$ ($x>0$).

$f'=2x-250/x^{2}=0\Rightarrow x^{3}=125\Rightarrow x=5$.

$f''(5)=2+500/125=6>0$ ⇒ relative minimum at $x=5$.

Relative minimum at $x=5$.

(2)Use the second-derivative test on $f(x)=x^{2}+\dfrac{250}{x}$ ($x>0$).

$f'=2x-250/x^{2}=0\Rightarrow x^{3}=125\Rightarrow x=5$.

$f''(5)=2+500/125=6>0$ ⇒ relative minimum at $x=5$.

Relative minimum at $x=5$.

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8–9. Proofs (guided)

8. First derivative test

9. Second derivative test

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10–11. Product optimizations

Book 10. $u+v=50$, maximize $P=uv$ ⇒ $u=v=25$, $P=625$ (table at endpoints $0$).

Book 11. Split $120$ to maximize $P=(120-x)x^{2}$ ⇒ parts $80$ and $40$.

Max product with fixed sum

P=u(S−u) on [0,S]; critical at S/2

(1)Among positive $u,v$ with $u+v=40$, maximize the product $P=uv$.

$P=u(40-u)=40u-u^{2}$ on $[0,40]$.

$P'=40-2u=0\Rightarrow u=20$, $v=20$.

Table / second-derivative test: max $P=400$.

$u=v=20$, max product $400$.

(2)Among positive $u,v$ with $u+v=60$, maximize the product $P=uv$.

$P=u(60-u)=60u-u^{2}$ on $[0,60]$.

$P'=60-2u=0\Rightarrow u=30$, $v=30$.

Table / second-derivative test: max $P=900$.

$u=v=30$, max product $900$.

Part × square of the other

P=(S−x)x²; critical at x=2S/3

(1)Split 120 into parts $x$ and 120-x$to maximize$P=(120-x)x^{2}$on$[0,120]$.

$P'=x(2\cdot120-3x)$; critical numbers $0$ and $80$.

Compare $P(0), P(80), P(120)$ — maximum at $x=80$.

Parts $80$ and $40$ (here 80 and 40 when integer).

(2)Split 90 into parts $x$ and 90-x$to maximize$P=(90-x)x^{2}$on$[0,90]$.

$P'=x(2\cdot90-3x)$; critical numbers $0$ and $60$.

Compare $P(0), P(60), P(90)$ — maximum at $x=60$.

Parts $60$ and $30$ (here 60 and 30 when integer).

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12. Poster with margins

Book. Area $18\,\mathrm{ft}^{2}$; vertical margins total $1$ ft, horizontal total $\tfrac{3}{2}$ ft. Printed $A=(x-1)(18/x-3/2)$; critical $x=2\sqrt{3}$; other side $3\sqrt{3}$ ft.

Fig. 14-10 — sheet vs printed rectangle

Maximize printed area

Express A(x), solve A′=0, confirm with A″ or uniqueness

(1)Poster area $18\,\mathrm{ft}^{2}$; top/bottom margins $9$ in ($3/4$ ft each side of height? book: $1$ ft total vertical margin), side margins $6$ in ($1/2$ ft each). Maximize printed area $A=(x-1)(18/x-3/2)$.

$A' = 18/x^{2}-3/2=0\Rightarrow x=2\sqrt{3}$.

$A''=-36/x^{3}<0$ at the critical number ⇒ relative max.

Unique critical number on $(0,\infty)$ ⇒ absolute max; other side $3\sqrt{3}$ ft.

Dimensions $2\sqrt{3}$ ft by $3\sqrt{3}$ ft.

(2)Sheet area $24$. Printed area $A=(x-1)(24/x-1)$ (1 ft margins each direction). Maximize $A$ for $x>1$.

$A'=24/x^{2}-1=0\Rightarrow x=\sqrt{24}$.

Second-derivative / unique critical number ⇒ absolute max of printed area.

Optimal width $x=\sqrt{24}$ (then height $24/x$).

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13. Two ships — nearest approach

Book. B starts $65$ mi east of A; B west $10$ mi/h, A south $15$ mi/h. Minimize $D$; nearest at $t=2$ h (11 a.m.), distance $15\sqrt{13}$ mi.

Fig. 14-11 — related rates / distance minimization

Nearest approach of two ships

Minimize D (or D²); unique critical time

(1)At $t=0$, ship B is 65 mi east of A. B sails west at 10 mi/h, A south at 15 mi/h. When are they nearest, and how near?

$D^{2}=(15t)^{2}+(65-10t)^{2}$; minimize $D$ via $dD/dt=0$.

$(325t-650)/D=0\Rightarrow t=2$ h (11 a.m. if start 9 a.m.).

First-derivative test $\{-,+\}$ + unique crit ⇒ abs min; $D=15\sqrt{13}$ mi.

Nearest at $t=2$ h; distance $15\sqrt{13}$ mi.

(2)B starts 65 mi east of A; B west at 12 mi/h, A south at 16 mi/h. Find $t$ minimizing separation.

$D^{2}=(16t)^{2}+(65-12t)^{2}$.

Set derivative of $D^{2}$ to $0$: $t=\dfrac{780}{400}$.

Unique positive critical number + first-derivative test ⇒ absolute minimum distance.

$t=\dfrac{39}{20}$ hours after start.

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14. Open / closed cylinder

Book. Volume $64$: (a) open can $r=h=4/\sqrt[3]{\pi}$; (b) closed can $h=2r=4\sqrt[3]{4/\pi}$.

Cylinder of fixed volume

Eliminate h from πr²h=V; minimize surface

(1)Closed cylinder, volume 64. Minimize $A=2\pi rh+2\pi r^{2}$.

$A=128/r+2\pi r^{2}$.

$A'=0\Rightarrow \pi r^{3}=32$ (book: $64\Rightarrow r=2\sqrt[3]{4/\pi}$, $h=2r$).

Unique positive critical number ⇒ absolute minimum surface.

Closed: $h=2r$ with $r=\sqrt[3]{64/(2\pi)}$.

(2)Closed cylinder, volume 125. Minimize $A=2\pi rh+2\pi r^{2}$.

$A=250/r+2\pi r^{2}$.

$A'=0\Rightarrow \pi r^{3}=62.5$ (book: $64\Rightarrow r=2\sqrt[3]{4/\pi}$, $h=2r$).

Unique positive critical number ⇒ absolute minimum surface.

Closed: $h=2r$ with $r=\sqrt[3]{125/(2\pi)}$.

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15. Profit and average cost

Book. Cost $\tfrac14 x^{2}+35x+25$, price $50-\tfrac12 x$. Max profit at $x=10$; average cost also has a relative min at $x=10$.

Max profit / min average cost

P′=0 and C′=0; second-derivative signs

(1)Price per item $60-x$, cost $x^{2}+10x+5$. Maximize profit $P=x(60-x)-(x^{2}+10x+5)$ for $x>0$.

$P=(60-10)x-2x^{2}-5$; $P'=50-4x=0\Rightarrow x=\dfrac{25}{2}$.

$P''=-4<0$ ⇒ relative (hence absolute, unique crit) maximum.

Produce $x=\dfrac{25}{2}$ items per day.

(2)Price per item $40-x$, cost $x^{2}+10x+5$. Maximize profit $P=x(40-x)-(x^{2}+10x+5)$ for $x>0$.

$P=(40-10)x-2x^{2}-5$; $P'=30-4x=0\Rightarrow x=\dfrac{15}{2}$.

$P''=-4<0$ ⇒ relative (hence absolute, unique crit) maximum.

Produce $x=\dfrac{15}{2}$ items per day.

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16. Locomotive — cost per mile

Book. Fuel $\propto v^{2}$ with \25$/h at$25$mi/h; other$$100$/h. Most economical speed$50$ mi/h.

Minimize cost per mile

C=(fuel+fixed)/v; solve C′=0 for v>0

(1)Fuel cost $\propto v^{2}$; \$25/h at$25mi/h. Other costs \100/h. Minimize cost per mile $C=(v/25+100)/v$.

$k=1/25$; $C=v/25+100/v$.

$C'=1/25-100/v^{2}=0\Rightarrow v=50$.

$C''>0$ at $50$ + unique crit on $(0,\infty)$ ⇒ abs min.

Most economical speed $50$ mi/h.

(2)Fuel cost $v^{2}/25$ dollars per hour; fixed costs $100$/h. Minimize $C=(v/25+100)/v$ for $v>0$.

$C'=1/25-100/v^{2}=0\Rightarrow v=5\sqrt{100}$.

Second-derivative test + uniqueness ⇒ absolute minimum cost/mile.

Speed $v=5\sqrt{100}$ mi/h.

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17. Row to shore, then walk

Book. $5$ mi offshore, destination $6$ mi along shore; row $2$ mi/h, walk $4$ mi/h. Land at $x=\tfrac{5}{3}\sqrt{3}\approx 2.89$ mi from A.

Fig. 14-12 — classical light-path / time minimization

Boat then walk

t(x)=row time + walk time; Snell-type critical equation

(1)Boat 5 mi offshore at P; destination B is 6 mi along shore from nearest point A. Row 2 mi/h, walk 4 mi/h. Land at C with $AC=x$ to minimize time.

$t=\sqrt{25+x^{2}}/2+(6-x)/4$.

$t'=0\Rightarrow 2x=\sqrt{25+x^{2}}\Rightarrow x=(5/3)\sqrt{3}\approx 2.89$.

Check endpoints / unique interior crit for absolute min time.

Land about $2.89$ mi from A toward B ($x=\tfrac{5}{3}\sqrt{3}$).

(2)Offshore 4 mi; walk 8 mi along shore. Row 2 mi/h, walk 4 mi/h. Set up $t(x)$ and the equation $t'=0$.

$t=\sqrt{16+x^{2}}/2+(8-x)/4$.

$t'=0\Leftrightarrow \dfrac{x}{2\sqrt{16+x^{2}}}=\dfrac{1}{4}$ (Snell form).

Solve $\dfrac{x}{2\sqrt{4^{2}+x^{2}}}=\dfrac{1}{4}$ for landing offset $x$.

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18. Fence along a river

Book. Fixed area $A=xy$, fencing $F=x+2y$ (no river side). Least fence when $x=2y$.

River fencing

Implicit differentiation / equate dy/dx conditions

(1)Rectangular field along a straight river (no fence on river side). Fixed area $A=xy$. Show least fencing $F=x+2y$ occurs when length $x$ is twice the width $y$.

From $A$ constant: $y+x\,dy/dx=0\Rightarrow dy/dx=-y/x$.

$dF/dx=1+2\,dy/dx=0\Rightarrow dy/dx=-1/2$.

Equate: $-y/x=-1/2\Rightarrow x=2y$. Second-derivative test confirms a minimum.

Least fence when length along the river is twice the width ($x=2y$).

(2)Rectangular field along a straight river (no fence on river side). Fixed area $A=xy$. Show least fencing $F=x+2y$ occurs when length $x$ is twice the width $y$.

From $A$ constant: $y+x\,dy/dx=0\Rightarrow dy/dx=-y/x$.

$dF/dx=1+2\,dy/dx=0\Rightarrow dy/dx=-1/2$.

Equate: $-y/x=-1/2\Rightarrow x=2y$. Second-derivative test confirms a minimum.

Least fence when length along the river is twice the width ($x=2y$).

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19. Cone about a sphere

Book. Sphere radius $8$; apex-to-center $y>8$. Similar triangles give $x/8=(y+8)/\sqrt{y^{2}-64}$, hence $V=\dfrac{64\pi(y+8)^{2}}{3(y-8)}$. Then $V'\propto(y+8)(y-24)/(y-8)^{2}$; the relevant critical number is $y=24$. Cone height $y+8=32$ in, base radius $x=8\sqrt{2}$ in. Unique critical number on $(8,\infty)$ plus a first-derivative sign change (or second-derivative test on $V$) shows the volume is minimized.

Fig. 14-13 — similar triangles for the circumscribed cone

Min-volume cone about a sphere

Similar triangles → V(y); solve V′=0 for y>R

(1)Right circular cone circumscribed about a sphere of radius $8$. With apex-to-center distance $y$ ($y>8$), show $V=\dfrac{64\pi(y+8)^{2}}{3(y-8)}$ and find the critical $y$.

Similar triangles: $x/8=(y+8)/\sqrt{y^{2}-64}$.

$V=\pi x^{2}(y+8)/3$ reduces to the displayed formula.

$V'\propto(y+8)(y-24)/(y-8)^{2}=0\Rightarrow y=24$ (with $y>8$).

Then height $y+8=32$, base radius $x=8\sqrt{2}$.

Critical $y=24$; cone height $32$, base radius $8\sqrt{2}$.

(2)Cone about a sphere of radius $4$. Using similar triangles with apex-to-center $y>4$, write $V(y)$ and the condition $V'=0$.

$x/4=(y+4)/\sqrt{y^{2}-16}$.

$V\propto (y+4)^{2}/(y-4)$; critical when $y=12$ (book pattern $y=24$ for $R=8$).

For radius $4$, expect critical $y=12$ (height $16$).

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20. Max-area rectangle in a parabola

Book. Inscribe $PBB'P'$ in $y^{2}=4px$ cut by $x=a$. With corner $(x,y)$, $A=2y(a-x)=2ay-y^{3}/(2p)$, so $A'=2a-3y^{2}/(2p)=0$ gives $y=\sqrt{4ap/3}$. Dimensions: height $2y=\dfrac{4}{3}\sqrt{3ap}$, depth $a-x=2a/3$. Since $A''=-3y/p<0$ and the critical number is unique, this is the maximum area.

Fig. 14-14 — inscribed rectangle in a parabolic segment

Rectangle in a parabolic segment

A=2y(a−x) with y²=4px; solve A′=0

(1)Parabola $y^{2}=4\cdot3\,x$ cut by $x=9$. Maximize inscribed rectangle area $A=2y(9-y^{2}/(4\cdot3))$.

$A'=2\cdot9-3y^{2}/(2\cdot3)=0\Rightarrow y=\sqrt{4\cdot9\cdot3/3}$.

Depth along $x$: $a-x=2a/3=6$.

$y=\sqrt{36}$; depth $6$.

(2)Parabola $y^{2}=4\cdot2\,x$ cut by $x=4$. Maximize inscribed rectangle area $A=2y(4-y^{2}/(4\cdot2))$.

$A'=2\cdot4-3y^{2}/(2\cdot2)=0\Rightarrow y=\sqrt{4\cdot4\cdot2/3}$.

Depth along $x$: $a-x=2a/3=2.6666666666666665$.

$y=\sqrt{10.666666666666666}$; depth $2.6666666666666665$.

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21. Max-volume cylinder in a sphere

Book. Cylinder inscribed in a sphere of radius $R$: $r^{2}+h^{2}=R^{2}$, $V=2\pi r^{2}h$. From $2r+2h\,dh/dr=0$ get $dh/dr=-r/h$; then $dV/dr=0$ yields $r^{2}=2h^{2}$, so $h=R/\sqrt{3}$ and full height $2h=2R/\sqrt{3}$.

Fig. 14-15 — inscribed cylinder (half-height h)

Cylinder in a sphere

Constraint r²+h²=R²; maximize V=2πr²h

(1)Right circular cylinder inscribed in a sphere of radius $R$. With half-height $h$ and base radius $r$ ($r^{2}+h^{2}=R^{2}$), maximize $V=2\pi r^{2}h$. Show the full height is $2R/\sqrt{3}$.

From $r^{2}+h^{2}=R^{2}$: $dh/dr=-r/h$.

$dV/dr=0\Rightarrow r^{2}=2h^{2}$, so $R^{2}=3h^{2}\Rightarrow h=R/\sqrt{3}$.

Full height $2h=2R/\sqrt{3}$. Second-derivative / uniqueness ⇒ max volume.

Cylinder height $2R/\sqrt{3}$.

(2)Sphere radius $9$. Inscribed cylinder: relate $r,h$ by $r^{2}+h^{2}=R^{2}$ and find the height $2h$ maximizing volume.

$V=2\pi r^{2}h$; at max, $r^{2}=2h^{2}$.

$h=9/\sqrt{3}$, full height $29/\sqrt{3}$.

Height $2\cdot9/\sqrt{3}$.

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22. Shortest beam over a wall

Book. Wall $10$ ft high, $8$ ft from the building. $L=\sqrt{(x+8)^{2}+y^{2}}$ with $y=10(x+8)/x$. Critical $x=2\sqrt[3]{100}$; shortest length $(\sqrt[3]{100}+4)^{3/2}$ ft. First-derivative test + Theorem 14.1 ⇒ absolute minimum.

Fig. 14-16 — similar triangles for the bracing beam

Shortest beam / ladder over a wall

Similar triangles → L(x); solve L′=0

(1)Brace a building with a beam over a parallel wall $10$ ft high and $8$ ft from the building. Minimize $L$ (Fig. 14-16 style).

Similar triangles: $y=10(x+8)/x$; $L=(x+8)\sqrt{x^{2}+100}/x$.

$L'\propto(x^{3}-800)/(\cdots)=0\Rightarrow x=2\sqrt[3]{100}$.

Shortest length $(\sqrt[3]{100}+4)^{3/2}$ ft; first-derivative test + uniqueness ⇒ absolute min.

$L=(\sqrt[3]{100}+4)^{3/2}$ ft at $x=2\sqrt[3]{100}$.

(2)Wall $8$ ft high is $27/8$ ft from a house. Shortest ladder leaning over the wall (same similar-triangles setup as the beam problem).

Critical when $x^{3}=8^{3}\cdot(27/8)$ in the book pattern, or equate $L'=0$.

Book answer: shortest ladder $15\tfrac{5}{8}$ ft.

Shortest ladder $15\tfrac{5}{8}$ ft.

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Mixed review

Max/min — mixed set

Domain extrema · derivative tests · applied optimization

(1)Among positive $u,v$ with $u+v=60$, maximize the product $P=uv$.

$P=u(60-u)=60u-u^{2}$ on $[0,60]$.

$P'=60-2u=0\Rightarrow u=30$, $v=30$.

Table / second-derivative test: max $P=900$.

$u=v=30$, max product $900$.

(2)Closed cylinder, volume 27. Minimize $A=2\pi rh+2\pi r^{2}$.

$A=54/r+2\pi r^{2}$.

$A'=0\Rightarrow \pi r^{3}=13.5$ (book: $64\Rightarrow r=2\sqrt[3]{4/\pi}$, $h=2r$).

Unique positive critical number ⇒ absolute minimum surface.

Closed: $h=2r$ with $r=\sqrt[3]{27/(2\pi)}$.

(3)B starts 40 mi east of A; B west at 8 mi/h, A south at 16 mi/h. Find $t$ minimizing separation.

$D^{2}=(16t)^{2}+(40-8t)^{2}$.

Set derivative of $D^{2}$ to $0$: $t=\dfrac{320}{320}$.

Unique positive critical number + first-derivative test ⇒ absolute minimum distance.

$t=1$ hours after start.

(4)Suppose $f'(x)=(x-(-3))(x-(-1))(x-(2))$ with three distinct real zeros. List critical numbers and the mono pattern from the sign chart of $f'$.

Critical numbers: $-3$, $-1$, $2$ (assume ordered $-3<-1<2$).

Cubic factor with positive leading coeff: $f'$ is $-$ left of leftmost root, then alternates.

Inc where $f'>0$, dec where $f'<0$.

Critical $-3,-1,2$; sign-chart $f'$ for mono intervals.

(5)Offshore 4 mi; walk 5 mi along shore. Row 2 mi/h, walk 4 mi/h. Set up $t(x)$ and the equation $t'=0$.

$t=\sqrt{16+x^{2}}/2+(5-x)/4$.

$t'=0\Leftrightarrow \dfrac{x}{2\sqrt{16+x^{2}}}=\dfrac{1}{4}$ (Snell form).

Solve $\dfrac{x}{2\sqrt{4^{2}+x^{2}}}=\dfrac{1}{4}$ for landing offset $x$.

(6)Fuel cost $v^{2}/25$ dollars per hour; fixed costs $120$/h. Minimize $C=(v/25+120)/v$ for $v>0$.

$C'=1/25-120/v^{2}=0\Rightarrow v=5\sqrt{120}$.

Second-derivative test + uniqueness ⇒ absolute minimum cost/mile.

Speed $v=5\sqrt{120}$ mi/h.

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Bonus · Supplementary problems

23(f–i). Classify relative extrema

Book answers.

• $(x^{2}-4)^{2}$: max $16$ at $0$; min $0$ at $\pm 2$.
• $(x-4)^{4}(x+3)^{3}$: max $6912$ at $0$; min $0$ at $4$; neither at $-3$.
• $x^{3}+48/x$: max $-32$ at $-2$; min $32$ at $2$.
• $(x-1)^{1/3}(x+2)^{2/3}$: max $0$ at $-2$; min $-\sqrt[3]{4}$ at $0$; neither at $1$.

Supp: classify relative extrema

Factor f′ / first-derivative test (incl. undefined points)

(1)Classify relative extrema of $f(x)=(x-1)^{1/3}(x+2)^{2/3}$.

Critical where derivative vanishes or is undefined: $-2,\,0,\,1$.

Rel max $0$ at $-2$; rel min $-\sqrt[3]{4}$ at $0$; neither at $1$.

Max $0$ at $-2$; min $-\sqrt[3]{4}$ at $0$; neither at $1$.

(2)Classify relative extrema of $f(x)=(x-1)^{1/3}(x+2)^{2/3}$.

Critical where derivative vanishes or is undefined: $-2,\,0,\,1$.

Rel max $0$ at $-2$; rel min $-\sqrt[3]{4}$ at $0$; neither at $1$.

Max $0$ at $-2$; min $-\sqrt[3]{4}$ at $0$; neither at $1$.

(3)Classify relative extrema of $f(x)=(x-1)^{1/3}(x+2)^{2/3}$.

Critical where derivative vanishes or is undefined: $-2,\,0,\,1$.

Rel max $0$ at $-2$; rel min $-\sqrt[3]{4}$ at $0$; neither at $1$.

Max $0$ at $-2$; min $-\sqrt[3]{4}$ at $0$; neither at $1$.

Problem 24

Revisit Problem 23(a–f) with the second-derivative test where $f'(c)=0$ and $f''(c)\neq 0$. If $f''(c)=0$ or $f'$ is undefined, fall back to the first-derivative test.

25. Sum of squared deviations

Claim. $y=\sum_{i=1}^{n}(a_i-x)^{2}$ has an absolute minimum at the arithmetic mean $x=(\sum a_i)/n$.

Supp: mean minimizes Σ(aᵢ−x)²

y′=0 ⇒ mean; y″=2n>0 ⇒ absolute min

(1)Show $y=\sum_{i=1}^{n}(a_i-x)^{2}$ has an absolute min at the mean. Example: a_{i}=6, a_{i}=3, a_{i}=2.

$y'= -2\sum(a_i-x)=0\Rightarrow x=(\sum a_i)/n$.

$y''=2n>0$ ⇒ absolute minimum at the arithmetic mean.

Absolute min at $x=\dfrac{11}{3}$.

(2)Show $y=\sum_{i=1}^{n}(a_i-x)^{2}$ has an absolute min at the mean. Example: a_{i}=8, a_{i}=1, a_{i}=1, a_{i}=2.

$y'= -2\sum(a_i-x)=0\Rightarrow x=(\sum a_i)/n$.

$y''=2n>0$ ⇒ absolute minimum at the arithmetic mean.

Absolute min at $x=3$.

26. Absolute extrema on an interval

Book samples. $-x^{2}$ on $(-2,2)$; $(x-3)^{2}$ on $[0,4]$; $\sqrt{25-4x^{2}}$ on $[-2,2]$; $\sqrt{x-4}$ on $[4,29]$.

Supp: absolute on an interval

Compare critical values with endpoints (watch open intervals)

(1)Absolute extrema of $y=-x^{2}$ on $(-2,2)$.

Open interval: max $0$ at $x=0$; no absolute min (approaches $-4$ at ends but ends excluded).

Abs max $0$ at $0$; no abs min on the open interval.

(2)Absolute extrema of $y=\sqrt{x-4}$ on $[4,29]$.

Increasing: min $0$ at $4$; max $5$ at $29$.

Max $5$ at $29$; min $0$ at $4$.

(3)Absolute extrema of $y=\sqrt{x-4}$ on $[4,29]$.

Increasing: min $0$ at $4$; max $5$ at $29$.

Max $5$ at $29$; min $0$ at $4$.

27–28. Two positive numbers

Book 27. Sum $20$: (a) max product $10,10$; (b) min sum of squares $10,10$; (c) max $x^{2}(20-x)^{3}$ → $8,12$.

Book 28. Product $16$: (a) min sum $4,4$; (b) min $x+y^{2}$ → $8,2$.

Supp: two-number optimizations

Fixed sum or product — one-variable calculus

(1)Positive numbers sum to $20$. Maximize product; minimize sum of squares; maximize (one)$^{2}$(other)$^{3}$.

(a) $10,10$. (b) $10,10$. (c) Critical from $P=x^{2}(20-x)^{3}$ ⇒ $8$ and $12$.

(a) $10,10$; (b) $10,10$; (c) $8,12$.

(2)Positive numbers sum to $20$. Maximize product; minimize sum of squares; maximize (one)$^{2}$(other)$^{3}$.

(a) $10,10$. (b) $10,10$. (c) Critical from $P=x^{2}(20-x)^{3}$ ⇒ $8$ and $12$.

(a) $10,10$; (b) $10,10$; (c) $8,12$.

29–46. Applied / geometry / physics

Classic setups (box cost, ladder, revenue, min-area triangle, distance to a curve, coil force, cell work, ellipse rectangle, cone/cylinder inscriptions, tent material, …).

#	Punchline
29	Box $20\times 20\times 16$
30	Ladder $15\tfrac{5}{8}$ ft
31	Order $65{,}000$
32	Line $4x+3y-24=0$
33	Point $(2\sqrt{3}/3,\,8/3)$
34	Distance $2\sqrt{2}$
37–38	$x=r/2$; $R=r$
40	Area $20\sqrt{2}\times 15\sqrt{2}$
41	Cone $R=\tfrac{2}{3}r\sqrt{2}$
42	Cyl in cone: $2r/3$ (vol), $r/2$ (lateral)
43	Tent $h=r\sqrt{2}$
45	Sphere $R=8$: $h=2r=8\sqrt{2}$
46	Total-area cylinder in cone: if $h>2r$, radius $\tfrac12 hr/(h-r)$

Supp: applied optimization

Boxes, ladders, triangles, inscriptions, physics work/force

(1)Force $F=kx/(x^{2}+r^{2})^{5/2}$. Show $F$ is greatest when $x=r/2$.

$F'=0\Rightarrow$ numerator factor $r^{2}-4x^{2}=0$ (for $x>0$) $\Rightarrow x=r/2$.

Max force at $x=\tfrac12 r$.

(2)Cylinder inscribed in a cone of base radius $r$: (a) max volume radius; (b) max lateral area radius.

(a) $R=\tfrac23 r$. (b) $R=\tfrac12 r$.

(a) $2r/3$; (b) $r/2$.

(3)On $y=4-x^{2}$ in Q1, point where tangent + axes make min-area triangle.

Tangent at $(x_0,4-x_0^{2})$; minimize triangle area.

Book: $(2\sqrt{3}/3,\,8/3)$.

$(2\sqrt{3}/3,\,8/3)$.

(4)Cylinder of max lateral area inscribed in a sphere of radius $8$. Dimensions?

Same geometry as Prob 21 with lateral $2\pi rh$ maximized.

Book: $h=2r=8\sqrt{2}$.

$h=2r=8\sqrt{2}$.

35–36. Implicit $y$-extrema / numeric absolute

Book 35. $2x^{2}-4xy+3y^{2}-8x+8y-1=0$: max $(5,3)$, min $(-1,-3)$.

Book 36 (GC). $f(x)=x^{5}-3x^{2}-8x-3$ on $[-1,2]$: max $\approx 1.191$ at $-0.866$; min $\approx -14.786$ at $1.338$.

Supp: implicit / numeric absolute

dy/dx=0 on an implicit curve, or GC critical scan

(1)Absolute max/min of $f(x)=x^{5}-3x^{2}-8x-3$ on $[-1,2]$ (three decimals; GC OK).

Solve $f'=0$ numerically on the interval; compare with endpoints.

Max $\approx 1.191$ at $x\approx -0.866$; min $\approx -14.786$ at $x\approx 1.338$.

Max $\approx 1.191$ at $-0.866$; min $\approx -14.786$ at $1.338$.

(2)Absolute max/min of $f(x)=x^{5}-3x^{2}-8x-3$ on $[-1,2]$ (three decimals; GC OK).

Solve $f'=0$ numerically on the interval; compare with endpoints.

Max $\approx 1.191$ at $x\approx -0.866$; min $\approx -14.786$ at $x\approx 1.338$.

Max $\approx 1.191$ at $-0.866$; min $\approx -14.786$ at $1.338$.

Bonus mixed

Supplementary mixed set

Problems 23–46 style

(1)Conical tent of fixed volume: show least material when height $h=r\sqrt{2}$.

$A=\pi r\sqrt{r^{2}+h^{2}}$ (book note uses $\pi(r^{2}+h^{2})$ form — lateral+floor variant).

With $V$ fixed, critical ratio $h=r\sqrt{2}$.

Least material when $h=r\sqrt{2}$.

(2)Show $y=\sum_{i=1}^{n}(a_i-x)^{2}$ has an absolute min at the mean. Example: a_{i}=1, a_{i}=4, a_{i}=7.

$y'= -2\sum(a_i-x)=0\Rightarrow x=(\sum a_i)/n$.

$y''=2n>0$ ⇒ absolute minimum at the arithmetic mean.

Absolute min at $x=4$.

(3)Classify relative extrema of $f(x)=(x-4)^{4}(x+3)^{3}$.

Critical at $4$, $-3$, and an interior factor zero $x=0$ (book).

Rel max $6912$ at $0$; rel min $0$ at $4$; neither at $-3$.

Max $6912$ at $0$; min $0$ at $4$; neither at $-3$.

(4)Absolute extrema of $y=\sqrt{25-4x^{2}}$ on $[-2,2]$.

Max $5$ at $0$; at $\pm 2$: $y=3$.

Max $5$ at $0$; min $3$ at $\pm 2$.

(5)Absolute max/min of $f(x)=x^{5}-3x^{2}-8x-3$ on $[-1,2]$ (three decimals; GC OK).

Solve $f'=0$ numerically on the interval; compare with endpoints.

Max $\approx 1.191$ at $x\approx -0.866$; min $\approx -14.786$ at $x\approx 1.338$.

Max $\approx 1.191$ at $-0.866$; min $\approx -14.786$ at $1.338$.

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Self-check

Quiz

Max/min practice review

1 / 6

A critical number is a domain point where…

Pick an answer to see if you are right.

Further Reading

• 15. Curve Sketching · Concavity · Symmetry — next theory chapter
• 14. Maximum and Minimum Values — theory, tests, absolute extrema
• 13b. Law of the Mean — Practice — prior chapter drills
• 13. Law of the Mean — monotone from $f'$