1 Maxima and Minima on finite closed intervals

1.1 Solution 1

$a + b = 3$

$N^2 = a^2 + 3^2$

$M^2 = a^2 + 3^2$

Total cables $ = N + M + b$

$= \sqrt{a^2 + 3^2} + \sqrt{a^2 + 3^2} + 3-a$

$f(a) = 2\sqrt{a^2 + 9} + 3 - a$

We need to find the local minima of the function.

Domain: $[0,3]$

Let's write GNU Octave code to solve this further:

pkg load symbolic
syms a
f = 2* sqrt(a^2 + 9) + 3 - a
df = simplify(diff(f,a))

solve(df == 0,a)

f = (sym)

            ________
           ╱  2
  -a + 2⋅╲╱  a  + 9  + 3
df = (sym)

      2⋅a
  ─────────── - 1
     ________
    ╱  2
  ╲╱  a  + 9
ans = (sym) √3

So the critical number is $\sqrt{3}$. Let's find the values of the function at the endpoints and the critical point:

fh = function_handle(f)
[fh(0), fh(sqrt(3)), fh(3)]

fh =

@(a) -a + 2 * sqrt (a .^ 2 + 9) + 3
ans =

   9.0000   8.1962   8.4853

So the minimum value for the function occurs at $a = \sqrt{3}$ which is at point $(0, \sqrt{3})$

1.2 Solution 2

Cost to produce gallon of maple syrup = $30$ dollars.

If they charge x dollar per gallon, then they could sell $1000 - 10x$ gallons.

$30 \leq x \leq 1000$

Let's express in terms of function.

x = Rate in dollars. x is the selling price.

$f(x) = 1000 - 10x$

$f(x)$ represents gallons sold.

1.2.1 Solution (a)

To maximize the revenue, they should sell lots of gallons.

$f(x) = 1000 - 10x$

Total gallons sold = f(x)

Revenue = Selling price = $x.f(x) = x(1000 - 10x)$

Cost $= 30.f(x)$

Revenue = Cost + Profit

Let's write some GNU Octave code to find which value of $x$ will maximize revenue.

$r(x) = x(1000 - 10x)$

clear
pkg load symbolic
syms x
r = x*(1000 - (10*x))
dr = simplify(diff(r,x))

solve(dr == 0, x)

r = (sym) x⋅(-10⋅x + 1000)
dr = (sym) -20⋅x + 1000
ans = (sym) 50

Let's fine the value of $x$ at which the function $r(x)$ is maximized:

rh = function_handle(r)
[rh(30), rh(50), rh(1000)]

rh =

@(x) x .* (-10 * x + 1000)
ans =

     21000     25000  -9000000

So they should set the price to $50 to maximize their revenue.

1.2.2 Solution (b)

p = x*(1000 - (10*x)) - 30*(1000-(10*x))
dp = simplify(diff(p, x))

solve(dp == 0, x)

p = (sym) x⋅(-10⋅x + 1000) + 300⋅x - 30000
dp = (sym) -20⋅x + 1300
ans = (sym) 65

Now let's see if $x$ is maximum at value $65$:

ph = function_handle(p)
[ph(30), ph(65), ph(1000)]

ph =

@(x) x .* (-10 * x + 1000) + 300 * x - 30000
ans =

         0     12250  -8730000

So the price should be set at $64 per barrel for the profit to be maximized:

1.3 Solution 3

Total area = 600 square inches

Margin at sides and top = 1 inch

Margin at bottom = 2 inches

Poster is rectangular.

Objective: Find dimension of poster to maximize printed area.

$a.b = 600$

Printed area $= (a-2)(b-3)$

$f(a) = (a-2)(\dfrac{600}{a} - 3)$

Let's write some GNU Octave code to find which value of $a$ at which the function attains it's maximum value.

clear
pkg load symbolic
syms a
f = (a-2)*(600/a - 3)

df = simplify(diff(f,a))
solve(df == 0, a)

f = (sym)

  ⎛     600⎞
  ⎜-3 + ───⎟⋅(a - 2)
  ⎝      a ⎠
df = (sym)

       1200
  -3 + ────
         2
        a
ans = (sym 2×1 matrix)

  ⎡-20⎤
  ⎢   ⎥
  ⎣20 ⎦

Since we know that the dimension cannot be negative, let's check if $20$ makes the function maximum:

fh = function_handle(f)
[fh(0), fh(20), fh(600)]

fh =

@(a) (-3 + 600 ./ a) .* (a - 2)
warning: division by zero
warning: called from
    eval>@<anonymous> at line 1 column 19
ans =

   -Inf    486  -1196

So we know that at $20$ the function achieves it's maximum value. The other dimension is $30$.

1.4 Solution 4

Solution described here

1.5 Solution 5

Permiter of window = P

Let's assume $l$ be the length of rectangle and $w$ be the width of the rectangle.

Area of rectangle $= l * w$

Radius of circle $= \dfrac{w}{2}$

Area of circle $= \dfrac{\pi * (\dfrac{w}{2})^2}{2} = \dfrac{\pi}{2} * \dfrac{w^2}{4} = \dfrac{w^2 \pi}{8}$

Area of norman window $= l.w + \dfrac{w^2 * \pi}{8}$

Permiter of norman window $= l + l + w + \pi . \dfrac{w}{2}$

$P = 2l + w(\dfrac{\pi}{2} + 1)$

Assuming that the permiter is constant, we need to find dimensions to maximize the are.

$P = 2l + w(\dfrac{\pi}{2} + 1)$

$l = \dfrac{P}{2} - \dfrac{2}{2}*(\dfrac{\pi}{2} + 1)$

$A = l.w + \dfrac{w^2 * \pi}{8}$

$f(w) = \dfrac{w^2\pi}{8} + w(\dfrac{P}{2} - \dfrac{w}{2}(\dfrac{\pi}{2} + 1))$

$f(w) = \dfrac{w^2\pi}{8} + w(\dfrac{P}{2} - (\dfrac{w\pi}{4} + \dfrac{w}{2}))$

$f(w) = \dfrac{w^2\pi}{8} + w(\dfrac{P}{2} - (\dfrac{w\pi + 2w}{4}))$

$= \dfrac{w^2\pi}{8} + w(\dfrac{P}{2} - (\dfrac{w(\pi + 2)}{4}))$

Let's write some GNU Octave code to solve it:

clear
pkg load symbolic
syms w P PI # using Q for pi, P for permiter, w for width
f = (w^2* PI)/8 + w*(P/2 - ((w*(PI + 2))/4))

df = simplify(diff(f,w))
solve(df == 0, w)

octave> octave> f = (sym)

      2
  PI⋅w      ⎛P   w⋅(PI + 2)⎞
  ───── + w⋅⎜─ - ──────────⎟
    8       ⎝2       4     ⎠
octave> df = (sym)

  P   PI⋅w
  ─ - ──── - w
  2    4
ans = (sym)

   2⋅P
  ──────
  PI + 4

Let's check if at $\dfrac{2.P}{\pi + 4}$, the function assumes it's maximum values. Let's take double derivate:

dff = simplify(diff(df, w))
dffh = function_handle(dff)
[dffh(pi)]

dff = (sym)

    PI
  - ── - 1
    4
dffh =

@(PI) -PI / 4 - 1
ans = -1.7854

Since the value is less than $0$, we know from second derivate test that the function achieves it's local maximum at that value. Now let's calculate what would be the length:

clear
pkg load symbolic
syms P W PI
l = P/2 - ((W/2)*(PI/2 + 1))

lh = function_handle(l)
[simplify(lh(P, PI, (2*P)/(PI + 4)))]

octave> octave> l = (sym)

        ⎛PI    ⎞
      W⋅⎜── + 1⎟
  P     ⎝2     ⎠
  ─ - ──────────
  2       2
octave> lh =

@(P, PI, W) P / 2 - W .* (PI / 2 + 1) / 2
ans = (sym)

    P
  ──────
  PI + 4

So the width of the rectangle is $\dfrac{2P}{\pi + 4}$ and the length of the windows is $\dfrac{P}{\pi + 4}$.

The height of the windows will be the sum of the length and radius of the circle. Let's compute that:

clear
pkg load symbolic
syms P PI
h = P/(PI + 4) + (P)/(PI + 4) # radius is same as half of the width of the window

octave> octave> h = (sym)

   2⋅P
  ──────
  PI + 4

And the height is same as the width of the windows.

1.6 Solution 6

Volume of cone using slant height $= \dfrac{1}{3} \pi r^2 \sqrt{l^2 - r^2}$

We know that $l = 3$

The range of radius can be $(0,l]$

Let's write some GNU Octave code to solve it:

clear
pkg load symbolic
syms r
f = 1/3 * pi * r^2 * sqrt(9 - r^2)

df = simplify(diff(f,r))
solve(df == 0, r)

octave> octave> warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 379 column 13
    mtimes at line 63 column 5
f = (sym)

          ________
     2   ╱      2
  π⋅r ⋅╲╱  9 - r
  ────────────────
         3
octave> df = (sym)

      ⎛     2⎞
  π⋅r⋅⎝6 - r ⎠
  ────────────
     ________
    ╱      2
  ╲╱  9 - r
ans = (sym 3×1 matrix)

  ⎡ 0 ⎤
  ⎢   ⎥
  ⎢-√6⎥
  ⎢   ⎥
  ⎣√6 ⎦

We know that the radius cannot be zero or negative. So let's check the critical number $\sqrt{6}$ to see if the function achives local maximum at that number.

fh = function_handle(f)
[fh(0), fh(sqrt(6)), fh(3)]

fh =

@(r) pi * r .^ 2 .* sqrt (9 - r .^ 2) / 3
ans =

    0.00000   10.88280    0.00000

From exterme value theorem, we can say that the function achieves it's local maximum when the radius is $\sqrt{6}$. Let's find the volume of the cup at that radius:

fh = function_handle(f)
[fh(sqrt(6))]

fh =

@(r) pi * r .^ 2 .* sqrt (9 - r .^ 2) / 3
ans =  10.883

So volume of 10.833 is the largest possible volume of the cup.

Reference: https://www.mathsisfun.com/geometry/cone.html

1.7 Solution 7

Way to go = Row boat + Walk to house

Total time taken $= \dfrac{y}{3} + \dfrac{100-x}{5}$

$y^2 = x^2 + 60^2$

$y = \sqrt{x^2 + 60^2}$

$f(x) = \dfrac{\sqrt{x^2 + 60^2}}{3} + \dfrac{100-x}{5}$

Let's write some GNU Octave code to solve it:

clear
pkg load symbolic
syms x
f = sqrt(x^2 + 60^2)/3 + (100-x)/5

df = simplify(diff(f,x))
solve(df == 0, x)

octave> octave> f = (sym)

           ___________
          ╱  2
    x   ╲╱  x  + 3600
  - ─ + ────────────── + 20
    5         3
octave> df = (sym)

         x           1
  ──────────────── - ─
       ___________   5
      ╱  2
  3⋅╲╱  x  + 3600
ans = (sym) 45

Now let's take double derivative of it:

dff = simplify(diff(df, x))
dffh = function_handle(dff)
[dffh(45)]

dff = (sym)

       1200
  ──────────────
             3/2
  ⎛ 2       ⎞
  ⎝x  + 3600⎠
dffh =

@(x) 1200 ./ (x .^ 2 + 3600) .^ (3 / 2)
ans =  0.0028444

Since the value is $> 0$, from second derivative test we can conclude that the function attains it's local minima at $x = 45$.

So, the little red riding hood should row to the point 45 feet west of the point directly across the river.

Reference: Little Red Riding hood

1.8 Solution 8

Total fence distance = 200 feet

Let's assume width of fence be w and length of fence be l.

Area $= w.l$

Total fence distance $= w + w + l = 2w + l$

$2w + l = 200$

$f(w) = w.l$

$f(w) = w(200-2w)$

Let's write some GNU Octave code to solve it:

clear
pkg load symbolic
syms w
f = w*(200 - 2*w)

df = simplify(diff(f,w))
solve(df == 0, w)

octave> octave> f = (sym) w⋅(200 - 2⋅w)
octave> df = (sym) 200 - 4⋅w
ans = (sym) 50

Now let's confirm if at $w = 50$, the function attains it's local maximum value. The domain for $w$ is $[0,100]$

fh = function_handle(f)
[fh(0), fh(50), fh(100)]

fh =

@(w) w .* (200 - 2 * w)
ans =

      0   5000      0

So by Extreme value theorem, we can conclude that at $w = 50$, the function attains it's local maxima. Let's compute the length of the field:

clear
pkg load symbolic
syms w
l = 200 - (2*w)
lh = function_handle(l)
[lh(50)]

octave> octave> l = (sym) 200 - 2⋅w
lh =

@(w) 200 - 2 * w
ans =  100

So the dimensions of the field should be width of $50$ and length of $100$ feet.

1.9 Solution 9

Total fence distance = 200 feet

Constraint: Field must be atleast as wide as barn

1.9.1 Solution (a)

Barn width = 40 feet

Let the dimensions of the field be $(40+x)$ and $y$.

Area $= (40+x).y$

$y+y+x+40+x = 200$

$2y + 2x + 40 = 200$

$2(x+y) = 160$

$x+y = 80$

$y = 80 - x$

Areaa $= (40+x)y$

$= (40+x)(80-x)$

Let's write some GNU Octave code to solve it:

clear
pkg load symbolic
syms x
f(x) = (40+x)*(80 - x)

df = simplify(diff(f,x))
solve(df == 0, x)

octave> octave> f(x) = (symfun) (80 - x)⋅(x + 40)
octave> df = (sym) 40 - 2⋅x
ans = (sym) 20

Now we need to verify that at the critical number $60$, the function achieves it's local maximum. We can assume domain of x to be $[0, 80$. So let's use exterme value theorem to check if it's achieve it's local maximum:

fh = function_handle(f)
[fh(0), fh(20), fh(80)]

warning: FIXME: symfun! does that need special treatment?
warning: called from
    codegen>codegen_input_parser at line 187 column 7
    codegen at line 21 column 35
    function_handle at line 94 column 12
fh =

@(x) (80 - x) .* (x + 40)
ans =

   3200   3600      0

So with that we know that one of the dimension is $20$ feet. Along with barn's dimension added, it becomes $60$ feet. Let's find the other dimension too:

clear
pkg load symbolic
syms x
y = 80 - x

yh = function_handle(y)
[yh(20)]

octave> octave> y = (sym) 80 - x
octave> yh =

@(x) 80 - x
ans =  60

So the dimensions of the field to maximize the area should be $60$ by $60$ feet.

1.9.2 Solution (b)

Barn width $= 70$ feet

Let the dimension of the field be $(70+x)$ and $y$

Area $= (70+x)y$

$y + y + x + (70+x) = 200$

$2x + 2y = 130$

$x + y = \dfrac{130}{2} = 65$

Area $= f(x) = (70+x)y$

$= (70+x)(\dfrac{130}{2} - x)$

Let's write some GNU Octave code to solve it:

clear
pkg load symbolic
syms x
f = (70+x)*(130/2 - x)

df = simplify(diff(f,x))
solve(df == 0, x)

octave> octave> f = (sym) (65 - x)⋅(x + 70)
octave> df = (sym) -2⋅x - 5
ans = (sym) -5/2

Hmm.. The critical points are at negative number. So for the above function in it's domain we don't have positive critical numbers. And that indicates that one of the dimension is a constant $70$ alongside the barn. Let's compute the other dimension:

clear
pkg load symbolic
syms y
y = 65 - x
yh = function_handle(y)

[yh(0)]

octave> octave> error: 'x' undefined near line 1 column 10
yh =

@(y) y
octave> ans = 0

So the dimensions of the fields are $60$ by $70$ feet.

1.10 Solution 10

Total fence perimeter $= 200$ feet.

Monthly rent $= $3$ per square feet.

Monthly rent for advertisement $= 30$ dollars per feet.

Let the dimension of the field be $l$ and $w$.

Area $= l.w$

$2l + 2w = 200$

$l + w = 100$

Area $= f(l) = l.w = l(100-l)$

Total monthly income $= l(100-l).3 + 30.l$

Let's write some GNU Octave code to solve it:

clear
pkg load symbolic
syms l
f = l*(100-l)*3 + (30*l)

df = simplify(diff(f,l))
solve(df==0,l)

octave> octave> f = (sym) 3⋅l⋅(100 - l) + 30⋅l
octave> df = (sym) 330 - 6⋅l
ans = (sym) 55

Now let's find if the above critical number is where the function attains it's local maxima. The domain for the function is $[0,100]$:

fh = function_handle(f)
[fh(0), fh(55), fh(100)]

fh =

@(l) 3 * l .* (100 - l) + 30 * l
ans =

      0   9075   3000

Let's find the other dimension of the field:

clear
pkg load symbolic
syms l
w = 100 - l

wh = function_handle(w)
[wh(55)]

octave> octave> w = (sym) 100 - l
octave> wh =

@(l) 100 - l
ans =  45

So the dimensions of $55$ and $45$ feet of the field are best to maximize his monthly income.