Module 14: Curve analysis and optimization

Factor the derivative and test intervals

Here $f'=3x^2-12x+9=3(x-1)(x-3)$. Its signs are positive before $1$, negative between $1$ and $3$, and positive after $3$. Thus the function increases on $(-\infty,1)$ and $(3,\infty)$ and decreases on $(1,3)$. At $1$ it has a local maximum $f(1)=4$; at $3$ a local minimum $f(3)=0$. These are local, not absolute on $\R$.

A second-derivative sign chart

Differentiate twice: $f''=12x^2-24x=12x(x-2)$. It is positive on $(-\infty,0)$ and $(2,\infty)$, negative on $(0,2)$. The graph is continuous and concavity changes at both candidate inputs. Its inflection points are $\boxed{(0,0),(2,-16)}$.

A candidate without a sign change

For $x^4$, $f''=12x^2\ge0$ on both sides of zero. The graph stays concave up, so zero is not an inflection input. In fact, it is a local and absolute minimum. A derivative being zero does not by itself describe how its sign behaves nearby.

Three fenced sides, not four

Let $w$ be each side perpendicular to the river and $\ell$ the parallel fenced side. The constraint is $2w+\ell=60$, so $A=w(60-2w)$ on $0<w<30$. Then $A'=60-4w=0$ gives $w=15$, $\ell=30$. Since $A''=-4<0$ and the endpoint limiting areas vanish, the largest area is $\boxed{450\text{ m}^2}$ with dimensions $15$ m by $30$ m.

A geometric domain eliminates false candidates

The base dimensions are both $12-2x$ and height is $x$. Thus

The feasible nondegenerate interval is $(0,6)$. Only $x=2$ is an interior critical input. $V'$ is positive before $2$ and negative afterward until $6$, and limiting endpoint volumes are zero. The maximum is $\boxed{128\text{ cm}^3}$ from a box $8$ cm by $8$ cm by $2$ cm. The zero at $6$ describes a collapsed box, not a second useful design.

Minimize a monotone transform of distance

Squared distance is $D(x)=x^2+(2x+1)^2=5x^2+4x+1$. Since square root is increasing, minimizing $D$ minimizes distance. Solve $D'=10x+4=0$ to get $x=-2/5$, hence $y=1/5$. The positive second derivative $10$ or completed square $D=5(x+2/5)^2+1/5$ proves a global minimum. The point is $\boxed{(-2/5,1/5)}$, at distance $1/\sqrt5$.

Keep disconnected domain pieces separate

Write $f=1-1/(x+1)$, with domain $x\ne-1$. The only intercept is $(0,0)$; asymptotes are vertical $x=-1$ and horizontal $y=1$. Since $f'=1/(x+1)^2>0$, it increases on each domain interval $(-\infty,-1)$ and $(-1,\infty)$. It is not increasing across the missing point as a single interval claim. Because $f''=-2/(x+1)^3$, concavity is up on the left and down on the right. There is no point of the graph at $x=-1$, so it is not an inflection point.

Optimize under a volume constraint

From $\pi r^2h=54\pi$, $h=54/r^2$ with $r>0$. A closed cylinder has

Thus $S'=4\pi r-108\pi/r^2=4\pi(r^3-27)/r^2$. Its unique positive zero is $r=3$ cm, giving $h=6$ cm and $S=54\pi$ cm$^2$. The derivative changes from negative to positive. Also $S\to\infty$ as $r\to0^+$ or $r\to\infty$, so the minimum is global. A very narrow cylinder needs huge height; a very wide one needs huge lids.

Factor the smallest power, then read one bracket

Rewrite: $f=x^{4/3}-4x^{1/3}$. With $\tfrac43-1=\tfrac13$ and $\tfrac13-1=-\tfrac23$,

Critical numbers: $x=1$, where $f'=0$, and $x=0$, where $f'$ is undefined but $f(0)=0$ is defined. (The definition agrees that $f'(0)$ does not exist: $\dfrac{f(h)-f(0)}{h}=h^{-2/3}(h-4)$ falls without bound as $h\to0$.) Since $x^{-2/3}=1/(x^{1/3})^2>0$ for $x\ne0$, $f'$ has the sign of $x-1$: negative for $x<1$, on both sides of $0$, and positive for $x>1$. So $f$ decreases on $(-\infty,1)$ and increases on $(1,\infty)$, with a local (in fact absolute) minimum $\boxed{f(1)=-3}$. There is no extremum at $0$: $f$ keeps falling through it, along a vertical tangent, since $f'\to-\infty$ there.

Factor out the exponential, which is never zero

$f'=e^{-x}+x(-e^{-x})=e^{-x}(1-x)$ and $f''=-e^{-x}(1-x)+e^{-x}(-1)=e^{-x}(x-2)$. Because $e^{-x}>0$, the brackets decide every sign. $f$ increases on $(-\infty,1)$ and decreases on $(1,\infty)$, with maximum $\boxed{f(1)=1/e\approx0.368}$, absolute by the one-critical-point rule. It is concave down on $(-\infty,2)$ and up on $(2,\infty)$, with inflection point $\boxed{(2,\,2/e^2)\approx(2,\,0.271)}$.

Derivative on top, then the second derivative test

$f'=2x\ln x+x^2\cdot\tfrac1x=x(2\ln x+1)$. For $x>0$ it is zero when $\ln x=-\tfrac12$, at $x=e^{-1/2}\approx0.607$. Next, $f''=2\ln x+2+1=2\ln x+3$, and $f''(e^{-1/2})=-1+3=2>0$, so this is a local minimum, $\boxed{f(e^{-1/2})=-\tfrac{1}{2e}\approx-0.184}$. Concavity changes where $2\ln x+3=0$, at $x=e^{-3/2}\approx0.223$: the inflection point is $\boxed{\left(e^{-3/2},\,-\tfrac{3}{2e^3}\right)\approx(0.223,\,-0.075)}$, concave down to its left and up to its right.

One identity, then the Wheel

$f'=2\sin x\cos x=\sin(2x)$, positive on $(0,\tfrac\pi2)$ and negative on $(\tfrac\pi2,\pi)$: $f$ rises to its maximum $\boxed{f(\tfrac\pi2)=1}$ and falls back, with minimum $0$ at both endpoints. Next, $f''=2\cos(2x)$, positive on $(0,\tfrac\pi4)$ and $(\tfrac{3\pi}4,\pi)$ and negative between. So $f$ is concave up, then down, then up, with inflection points $\boxed{(\tfrac\pi4,\tfrac12)}$ and $\boxed{(\tfrac{3\pi}4,\tfrac12)}$.

When one test is silent, ask the other

$f'=4x^3-12x^2=4x^2(x-3)$, so the critical numbers are $0$ and $3$, and $f''=12x^2-24x$. At $3$: $f''(3)=108-72=36>0$, a local minimum, $\boxed{f(3)=-27}$. At $0$: $f''(0)=0$, so the second derivative test says nothing. First derivative test: near $0$, $4x^2>0$ and $x-3<0$, so $f'<0$ on both sides. $f$ keeps decreasing through $0$: no extremum there.

Symmetry first, then two quotient rules

$f(-x)=-f(x)$, so the graph is symmetric about the origin. The bottom is never zero, so there is no vertical asymptote, and $f\to0$ as $x\to\pm\infty$: the horizontal asymptote is $y=0$. Quotient rule: $f'=\dfrac{(x^2+1)-x(2x)}{(x^2+1)^2}=\dfrac{1-x^2}{(x^2+1)^2}$, so $f$ increases on $(-1,1)$ and decreases on $(-\infty,-1)$ and $(1,\infty)$, with maximum $(1,\tfrac12)$ and minimum $(-1,-\tfrac12)$, both absolute. Workbench for $f''$, with $u=1-x^2$ and $v=(x^2+1)^{-2}$: $u'=-2x$ and $v'=-4x(x^2+1)^{-3}$, so

It is negative on $(-\infty,-\sqrt3)$ and $(0,\sqrt3)$ and positive on $(-\sqrt3,0)$ and $(\sqrt3,\infty)$, giving inflection points $\boxed{(-\sqrt3,-\tfrac{\sqrt3}{4}),\ (0,0),\ (\sqrt3,\tfrac{\sqrt3}{4})}$.

Expensive lids make a taller box

Workbench. (1) Minimize the cost $C$ in cents, with base side $x$ and height $h$ in cm. (2) $C=2(2x^2)+1(4xh)$. (3) $x^2h=2000$, so $h=2000/x^2$ and $C(x)=4x^2+\dfrac{8000}{x}$ for $x>0$. (4) With $\dfrac{8000}{x}=8000x^{-1}$, $C'=8x-8000x^{-2}=8x-\dfrac{8000}{x^2}=\dfrac{8(x^3-1000)}{x^2}$, zero at $x=10$. (5) $C'<0$ before $10$ and $C'>0$ after, so this is the global minimum. The cheapest box is $\boxed{10\text{ cm}\times10\text{ cm}\times20\text{ cm}}$, costing $400+800=1200$ cents, $\boxed{\$12.00}$. Because the lids cost double, the best box is twice as tall as it is wide.

Minimize the square, report the root

The squared distance (M14.06 called it $D$; here it is $D^2$, so that $D$ is the distance itself) is $D^2=(x-3)^2+(\sqrt x)^2=(x-3)^2+x$ for $x\ge0$. Its derivative $2(x-3)+1$ is zero at $x=\tfrac52$, its second derivative is $2>0$, and it is the only critical point, so it is the global minimum (also $D^2(0)=9$ is larger). The closest point is $\boxed{\left(\tfrac52,\sqrt{\tfrac52}\right)\approx(2.5,\,1.581)}$, with $D^2=\tfrac14+\tfrac52=\tfrac{11}4$, so the distance is $\boxed{\tfrac{\sqrt{11}}{2}\approx1.658}$.

The critical point is the worst cut

If $x$ metres go to the square, its side is $x/4$; the circle’s circumference is $10-x$, so its radius is $(10-x)/(2\pi)$. Total area on $[0,10]$:

Setting $A'=0$ and multiplying by $8\pi$: $\pi x=4(10-x)$, so $x=\dfrac{40}{\pi+4}\approx5.60$ m. Since $A''=\tfrac18+\tfrac1{2\pi}>0$, this is the minimum: $A=\dfrac{25}{\pi+4}\approx\boxed{3.50\text{ m}^2}$, cutting $5.60$ m for the square. For the maximum compare the endpoints: $A(0)=\dfrac{25}{\pi}\approx7.96$ m$^2$ (all circle) and $A(10)=6.25$ m$^2$ (all square). The largest area comes from not cutting at all: bend the whole wire into a circle, $\boxed{\approx7.96\text{ m}^2}$. The circle encloses the most area for its length, so the maximum sits at an endpoint.

The inside angle doubles

$R'(\theta)=\dfrac{v^2}{g}\cos(2\theta)\cdot2$, which is zero on $(0,\tfrac\pi2)$ only at $2\theta=\tfrac\pi2$, that is $\theta=\tfrac\pi4$. The endpoints give $R=0$, so the maximum is at $\boxed{45^\circ}$, with $R_{\max}=\dfrac{v^2}{g}=\dfrac{400}{9.81}\approx\boxed{40.8\text{ m}}$. Half the maximum needs $\sin(2\theta)=\tfrac12$, so $2\theta=30^\circ$ or $150^\circ$: $\boxed{\theta=15^\circ\text{ or }75^\circ}$. These angles add to $90^\circ$; complementary angles always give equal ranges, because $\sin(2\theta)=\sin(180^\circ-2\theta)$.

Induced drag equals parasite drag

(a) Quotient rule: