Applied Optimization

Objective. Let $x$ be the length (m) of each of the two sides perpendicular to the river and $y$ the length of the side parallel to it. The area is

$A = xy .$

Constraint. Only three sides are fenced, and all the fencing is used:

$2x + y = 400 \quad\Longrightarrow\quad y = 400-2x .$

Reduce to one variable.

$A(x) = x(400-2x) = 400x - 2x^{2}.$

Domain. We need $x\ge0$ and $y=400-2x\ge0$, i.e. $x\le200$. So $I=[0,200]$. $A$ is a polynomial, hence continuous on this closed bounded interval, so by the Extreme Value Theorem a maximum exists and the Closed Interval Method applies.

Critical numbers. $A'(x)=400-4x$, which exists for every $x$, so the only critical number comes from

$400-4x=0 \quad\Longrightarrow\quad x=100 ,$

and $100$ lies in $(0,200)$.

Compare the candidates.

Conclusion. The maximum area is $\mathbf{20000\ \text{m}^{2}}$, attained with the two sides perpendicular to the river equal to $x=100$ m and the side parallel to the river equal to $y=400-200=200$ m. The optimal field is twice as long (along the river) as it is deep.

Independent check. $A''(x)=-4<0$ for every $x$, so $A$ is concave down everywhere and its unique stationary point must be the absolute maximum — the same conclusion by a different route. Numerically, $A(99)=99(202)=19998$ and $A(101)=101(198)=19998$, both less than $20000$.

Objective and constraint. Let the sides be $x$ and $y$ metres. The constraint is $xy=64$, so $y=\dfrac{64}{x}$, and the perimeter is

$P = 2x+2y = 2x + \frac{128}{x}.$

Domain. A side must have positive length, and $64/x$ is undefined at $x=0$, so $I=(0,\infty)$. This interval is open and unbounded: the Extreme Value Theorem does not apply, and there are no endpoints to evaluate, so the Closed Interval Method cannot be used.

Critical numbers.

$P'(x) = 2-\frac{128}{x^{2}} = \frac{2x^{2}-128}{x^{2}} = \frac{2(x-8)(x+8)}{x^{2}} .$

On $(0,\infty)$, $P'$ exists everywhere and vanishes only when $x=8$ (the root $x=-8$ is not in the domain).

Justification — either argument is complete on its own.

Sign of $P'$. For $0<x<8$ we have $x-8<0$ and $x+8>0$, so $P'(x)<0$; for $x>8$ both factors are positive, so $P'(x)>0$. Thus $P$ is strictly decreasing on $(0,8]$ and strictly increasing on $[8,\infty)$, and by the First Derivative Test for Absolute Extreme Values $P(8)$ is the absolute minimum on $(0,\infty)$.

Global concavity. $P''(x)=\dfrac{256}{x^{3}}>0$ for every $x>0$, so $P$ is concave up on the entire interval and the unique stationary point is the absolute minimum.

Answer. $x=8$ m and $y=\dfrac{64}{8}=8$ m: the garden is a square of side $8$ m, and the minimum amount of fencing is

$P(8)=2(8)+\frac{128}{8}=16+16=\mathbf{32\ \text{m}} .$

Check. $P(4)=8+32=40$, $P(8)=32$, $P(16)=32+8=40$ — consistent with a minimum at $x=8$. Note also that $P(x)\to\infty$ both as $x\to0^{+}$ and as $x\to\infty$, so there is no maximum: an area-$64$ rectangle can have arbitrarily large perimeter.

Objective. After the corners are removed the base is a square of side $12-2x$ and the height is $x$, so

$V(x) = x(12-2x)^{2}.$

Domain. We need $x\ge0$ and $12-2x\ge0$, so $I=[0,6]$. $V$ is a polynomial, hence continuous on this closed bounded interval: the Extreme Value Theorem guarantees a maximum and the Closed Interval Method applies.

Critical numbers. By the product rule and the chain rule,

$V'$ exists for all $x$, and $V'(x)=0$ when $x=6$ or $x=2$. The value $x=6$ is an endpoint of the domain, so the only interior critical number is $x=2$.

Compare the candidates.

Conclusion. Cut squares of side $\mathbf{x=2}$ cm; the maximum volume is $\mathbf{128\ \text{cm}^{3}}$, from a box measuring $8\times8\times2$ cm.

Check two ways. Expanding first gives $V(x)=4x^{3}-48x^{2}+144x$ and $V'(x)=12x^{2}-96x+144=12(x-2)(x-6)$, matching the factored derivative above. Numerically, $V(1.9)=1.9(8.2)^{2}=127.76$ and $V(2.1)=2.1(7.8)^{2}=127.76$, both below $128$.

Set-up. Let $x$ be the side of the square base and $h$ the height (cm).

Constraint. $x^{2}h = 32000$, so $h=\dfrac{32000}{x^{2}}$.

Objective. The box has a base and four sides but no top:

$S = x^{2}+4xh = x^{2}+4x\cdot\frac{32000}{x^{2}} = x^{2}+\frac{128000}{x} .$

Domain. $x>0$ (the expression $128000/x$ is undefined at $x=0$, and a base of side $0$ encloses nothing), so $I=(0,\infty)$ — open, so no Closed Interval Method.

Critical numbers.

$S'(x) = 2x-\frac{128000}{x^{2}} = \frac{2x^{3}-128000}{x^{2}} = \frac{2\left(x^{3}-64000\right)}{x^{2}} .$

$S'$ exists for every $x>0$, and $S'(x)=0$ exactly when $x^{3}=64000$, i.e. $x=40$.

Justification. $S''(x)=2+\dfrac{256000}{x^{3}}>0$ for every $x>0$, so $S$ is concave up on the whole of $(0,\infty)$; therefore the unique stationary point gives the absolute minimum. (Equivalently: $x^{3}-64000<0$ for $0<x<40$, so $S'<0$ there, and $x^{3}-64000>0$ for $x>40$, so $S'>0$ there — $S$ decreases then increases.)

Answer.

$h = \frac{32000}{40^{2}} = \frac{32000}{1600} = 20 ,$

so the box is $\mathbf{40\ \text{cm}\times40\ \text{cm}}$ at the base with height $\mathbf{20\ \text{cm}}$ — note $h=\tfrac12 x$. The minimum surface area is

$S(40)=40^{2}+\frac{128000}{40}=1600+3200=\mathbf{4800\ \text{cm}^{2}} .$

Check. Volume: $40^{2}\cdot20 = 32000$ ✓. Neighbouring values: $S(39)=1521+\dfrac{128000}{39}\approx1521+3282.05=4803.05$ and $S(41)=1681+\dfrac{128000}{41}\approx1681+3121.95=4802.95$, both larger than $4800$ ✓.

Objective. A general point of the parabola is $\left(x,x^{2}\right)$, and its distance to $(0,3)$ is

$d(x)=\sqrt{(x-0)^{2}+\left(x^{2}-3\right)^{2}} .$

Because $t\mapsto\sqrt{t}$ is strictly increasing on $[0,\infty)$, $d$ and $g=d^{2}$ attain their minimum at exactly the same $x$-values, so minimise

Domain. All real $x$ — the whole parabola is available.

Critical numbers.

$g'(x)=4x^{3}-10x = 2x\left(2x^{2}-5\right) ,$

which exists everywhere and vanishes at $x=0$ and $x=\pm\sqrt{\tfrac52}=\pm\dfrac{\sqrt{10}}{2}\approx\pm1.5811$.

Sign analysis (write $x_{0}=\tfrac{\sqrt{10}}{2}$):

So $g$ has local minima at $x=\pm x_{0}$ and a local maximum at $x=0$. Since $g$ decreases on $(-\infty,-x_{0}]$ and increases on $[x_{0},\infty)$ — and $g$ is even, so $g(-x_{0})=g(x_{0})$ — the value $g(\pm x_{0})$ is the absolute minimum of $g$ on the whole real line. (Consistently, $g(x)=x^{4}-5x^{2}+9\to+\infty$ as $x\to\pm\infty$.)

Value. With $x_{0}^{2}=\tfrac52$,

$g(\pm x_{0}) = \left(\frac52\right)^{2}-5\left(\frac52\right)+9 = \frac{25}{4}-\frac{25}{2}+9 = \frac{25-50+36}{4} = \frac{11}{4}.$

Hence the minimum distance is $d=\sqrt{\dfrac{11}{4}}=\dfrac{\sqrt{11}}{2}\approx1.658$.

Answer. There are two closest points, symmetric about the $y$-axis:

$\left(\frac{\sqrt{10}}{2},\ \frac{5}{2}\right)\quad\text{and}\quad\left(-\frac{\sqrt{10}}{2},\ \frac{5}{2}\right),\qquad \text{shortest distance } \frac{\sqrt{11}}{2}\approx1.66 .$

Warning and check. The vertex $(0,0)$ is a critical number but a local maximum of $g$ restricted to a neighbourhood of $0$; its distance to $(0,3)$ is $3$, far larger than $1.658$. Numerically, $g(1.5)=5.0625-11.25+9=2.8125$ and $g(1.6)=6.5536-12.8+9=2.7536$ and $g(1.7)=8.3521-14.45+9=2.9021$, all above $g(x_{0})=2.75$ ✓.

Set-up. Take the semicircle to be the upper half of $x^{2}+y^{2}=4$, i.e. $y=\sqrt{4-x^{2}}$ for $-2\le x\le 2$. By symmetry the optimal rectangle is symmetric about the $y$-axis, with vertices $(\pm x,0)$ and $\left(\pm x,\sqrt{4-x^{2}}\right)$ for some $x$ with $0\le x\le2$. Its width is $2x$ and its height is $\sqrt{4-x^{2}}$, so

$A(x)=2x\sqrt{4-x^{2}},\qquad 0\le x\le 2 .$

$A$ is continuous on the closed bounded interval $[0,2]$, so the Extreme Value Theorem applies and the Closed Interval Method will settle the question.

Critical numbers. For $0<x<2$, by the product and chain rules,

$A'(x)=0$ requires $8-4x^{2}=0$, i.e. $x^{2}=2$, so $x=\sqrt2$ (the negative root is outside the domain). $A'$ also fails to exist at $x=2$, but that is an endpoint, which the Closed Interval Method tests anyway.

Compare the candidates.

Answer. The largest rectangle has width $2\sqrt2\approx2.83$ and height $\sqrt{4-2}=\sqrt2\approx1.41$ — it is exactly twice as wide as it is tall — and its area is $\mathbf{4}$ square units.

Check. $A(1.3)=2.6\sqrt{2.31}\approx2.6(1.5199)=3.952$ and $A(1.5)=3\sqrt{1.75}\approx3(1.3229)=3.969$, both less than $4$ ✓.

Worth remembering. For a semicircle of radius $R$ the same computation gives $x=\dfrac{R}{\sqrt2}$, height $\dfrac{R}{\sqrt2}$, width $R\sqrt2$, and maximum area $R^{2}$. With $R=2$ this returns $4$ ✓.

Set-up. Let $r$ be the radius of the semicircle, so the rectangle has width $2r$, and let $h$ be the height of the rectangle (metres).

Constraint. The boundary consists of the bottom edge $2r$, the two vertical sides $2h$, and the semicircular arc of length $\pi r$ (the top edge of the rectangle is interior, not part of the perimeter):

$2r+2h+\pi r = 10 \quad\Longrightarrow\quad h=\frac{10-(2+\pi)r}{2}.$

Objective.

Domain. We need $r\ge0$ and $h\ge0$, i.e. $(2+\pi)r\le10$, so

$0\le r\le \frac{10}{2+\pi}\approx1.9449 .$

$A$ is a polynomial in $r$, hence continuous on this closed bounded interval: the Closed Interval Method applies.

Critical numbers.

$A'(r) = 10-4r-\pi r = 10-(4+\pi)r ,$

which exists everywhere and vanishes at

$r^{*}=\frac{10}{4+\pi}\approx1.4002 ,$

and indeed $\dfrac{10}{4+\pi}<\dfrac{10}{2+\pi}$, so $r^{*}$ is interior to the domain.

Compare the candidates.

$A(0)=0 .$

At $r=\dfrac{10}{2+\pi}$ the rectangle collapses ($h=0$) and the window is a bare semicircle:

$A\left(\frac{10}{2+\pi}\right)=\frac{\pi}{2}\left(\frac{10}{2+\pi}\right)^{2}=\frac{50\pi}{(2+\pi)^{2}}\approx\frac{157.08}{26.436}\approx5.941 .$

At $r=r^{*}$, using $A(r)=10r-\dfrac{4+\pi}{2}r^{2}$,

The height. With $r=r^{*}$,

$10-(2+\pi)r^{*}=\frac{10(4+\pi)-10(2+\pi)}{4+\pi}=\frac{20}{4+\pi},$

so $h=\dfrac{10}{4+\pi}=r^{*}$.

Answer. $r=h=\dfrac{10}{4+\pi}\approx1.40$ m — the rectangle's height equals the radius, so the rectangle is twice as wide as it is tall — and the maximum area is

$A_{\max}=\frac{50}{4+\pi}\approx 7.00\ \text{m}^{2}.$

Justification, checked twice. The Closed Interval Method already decides it: $7.0012>5.941>0$. Independently, $A''(r)=-(4+\pi)<0$ for every $r$, so $A$ is concave down on the whole interval and its unique stationary point is the absolute maximum.

Sanity check of the perimeter. $2r+2h+\pi r = 4r+\pi r = (4+\pi)r^{*} = (4+\pi)\cdot\dfrac{10}{4+\pi}=10$ ✓.

Set-up. Let $w$ be the width, so the length is $2w$, and let $h$ be the height (all in metres).

Constraint. $V=(2w)(w)h=2w^{2}h=10$, so

$h=\frac{5}{w^{2}} .$

Objective. Base area $=2w^{2}$, costing $10\left(2w^{2}\right)=20w^{2}$ dollars. The four sides consist of two rectangles of area $wh$ and two of area $2wh$, a total area of $2wh+4wh=6wh$, costing $6(6wh)=36wh$ dollars. Hence

$C = 20w^{2}+36wh = 20w^{2}+36w\cdot\frac{5}{w^{2}} = 20w^{2}+\frac{180}{w},\qquad w>0 .$

Domain $(0,\infty)$ — open, so the Closed Interval Method is unavailable.

Critical numbers.

$C'(w) = 40w-\frac{180}{w^{2}} = \frac{40w^{3}-180}{w^{2}} .$

This exists for all $w>0$ and vanishes when

$40w^{3}=180 \quad\Longrightarrow\quad w^{3}=\frac{9}{2} \quad\Longrightarrow\quad w=\left(\frac92\right)^{1/3}\approx1.6510 .$

Justification. $C''(w)=40+\dfrac{360}{w^{3}}>0$ for every $w>0$, so $C$ is concave up on the whole domain and the unique stationary point is the absolute minimum. (Also $C(w)\to\infty$ as $w\to0^{+}$ and as $w\to\infty$, so there is no maximum.)

Minimum cost. Using $w^{3}=\tfrac92$ to avoid rounding,

$\frac{180}{w} = \frac{180w^{2}}{w^{3}} = \frac{180w^{2}}{9/2} = 40w^{2},$

so

$C_{\min}=20w^{2}+40w^{2}=60w^{2}=60\left(\frac92\right)^{2/3}\approx 60(2.72568)\approx 163.54 .$

Answer. Width $w=\left(\tfrac92\right)^{1/3}\approx1.651$ m, length $2w\approx3.302$ m, height $h=\dfrac{5}{w^{2}}\approx\dfrac{5}{2.7257}\approx1.834$ m, and the minimum cost is about $\mathbf{163.54}$ dollars.