CLP-1 Differential Calculus

A closed rectangular container with a square base is to be made from two different materials. The material for the base costs $5 per square meter, while the material for the other five sides costs $1 per square meter. Find the dimensions of the container which has the largest possible volume if the total cost of materials is $72.

Solution We can follow the steps we outlined above to find the solution.

• We need to determine the area of the two types of materials used and the corresponding total cost.

• Draw a picture of the box.

  

  The more useful picture is the unfolded box on the right.

• In the picture we have already introduced two variables. The square base has side-length \(b\) metres and it has height \(h\) metres. Let the area of the base be \(A_b\) and the area of the other fives sides be \(A_s\) (both in \(m^2\)), and the total cost be \(C\) (in dollars). Finally let the volume enclosed be \(V m^3\text{.}\)
• Some simple geometry tells us that
  \begin{align*} A_b &= b^2\\ A_s &= 4 bh + b^2\\ V &= b^2h\\ C &= 5 \cdot A_b + 1\cdot A_s = 5b^2+4bh+b^2 = 6b^2+4bh. \end{align*}
• To eliminate one of the variables we use the fact that the total cost is $72.
  \begin{align*} C &= 6b^2+4bh = 72 & \text{rearrange}\\ 4bh &= 72-6b^2 & \text{isolate } h\\ h &= \frac{72-6b^2}{4b} = \frac{3}{2} \cdot \frac{12-b^2}{b} \end{align*}
  Substituting this into the volume gives
  \begin{align*} V&= b^2 h = \frac{3b}{2} (12-b^2) = 18b - \frac{3}{2} b^3 \end{align*}
  Now note that since \(b\) is a length it cannot be negative, so \(b \geq 0\text{.}\) Further since volume cannot be negative, we must also have
  \begin{gather*} 12-b^2 \geq 0 \end{gather*}
  and so \(b \leq \sqrt{12}\text{.}\)
• Now we can apply Corollary 3.5.13 on the above expression for the volume with \(0 \leq b \leq \sqrt{12}\text{.}\) The endpoints give:
  \begin{align*} V(0) &= 0\\ V(\sqrt{12}) &= 0 \end{align*}
  The derivative is
  \begin{align*} V'(b) &= 18 - \frac{9b^2}{2} \end{align*}
  Since this is a polynomial there are no singular points. However we can solve \(V'(b) = 0\) to find critical points:
  \begin{align*} 18 - \frac{9b^2}{2} &= 0 & \text{divide by 9 and multiply by 2}\\ 4 - b^2 &= 0 \end{align*}
  Hence \(b = \pm 2\text{.}\) Thus the only critical point in the domain is \(b=2\text{.}\) The corresponding volume is
  \begin{align*} V(2) &= 18\times2 - \frac{3}{2} \times 2^3\\ &= 36 - 12 = 24. \end{align*}
  So by Corollary 3.5.13, the maximum volume is when 24 when \(b=2\) and
  \begin{align*} h &= \frac{3}{2} \cdot \frac{12-b^2}{b} = \frac{3}{2} \frac{12-4}{2} = 6. \end{align*}
• All our quantities make sense; lengths, areas and volumes are all non-negative.

• Checking the question again, we see that we are asked for the dimensions of the container (rather than its volume) so we can answer with

  The container with dimensions \(2 \times 2 \times 6m\) will be the largest possible.

A rectangular sheet of cardboard is 6 inches by 9 inches. Four identical squares are cut from the corners of the cardboard, as shown in the figure below, and the remaining piece is folded into an open rectangular box. What should the size of the cut out squares be in order to maximize the volume of the box?

Solution This one is quite similar to the previous one, so we perhaps don't need to go into so much detail.

• After reading carefully we produce the following picture:

  
• Let the height of the box be \(x\) inches, and the base be \(\ell \times w\) inches. The volume of the box is then \(V\) cubic inches.
• Some simple geometry tells us that \(\ell = 9-2x, w=6-2x\) and so
  \begin{align*} V &= x(9-2x)(6-2x) \text{cubic inches}\\ &= 54x-30x^2+4x^3. \end{align*}
  Notice that since all lengths must be non-negative, we must have
  \begin{gather*} x,\ell,w \geq 0 \end{gather*}
  and so \(0 \leq x \leq 3\) (if \(x \gt 3\) then \(w \lt 0\)).

• We can now apply Corollary 3.5.13. First the endpoints of the interval give

  \begin{align*} V(0) &= 0 & V(3) &= 0 \end{align*}

  The derivative is

  \begin{align*} V'(x) &= 54 - 60x +12x^2\\ &= 6(9-10x+2x^2) \end{align*}

  Since this is a polynomial there are no singular points. To find critical points we solve \(V'(x) = 0\) to get

  \begin{align*} x_\pm &= \frac{10 \pm \sqrt{100 - 4\times2\times9}}{4}\\ &= \frac{10 \pm \sqrt{28}}{4} = \frac{10 \pm 2\sqrt{7}}{4} = \frac{5 \pm \sqrt{7}}{2} \end{align*}

  We can then use a calculator to approximate

  \begin{align*} x_+ &\approx 3.82 & x_- &\approx 1.18. \end{align*}

  So \(x_-\) is inside the domain, while \(x_+\) lies outside.

  Alternatively ⁷Say if we do not have a calculator to hand, or your instructor insists that the problem be done without one. , we can bound \(x_\pm\) by first noting that \(2 \leq \sqrt{7} \leq 3\text{.}\) From this we know that

  \begin{align*} 1=\frac{5-3}{2} & \leq x_- = \frac{5 - \sqrt{7}}{2} \leq \frac{5-2}{2} = 1.5\\ 3.5=\frac{5+2}{2} & \leq x_+ = \frac{5 + \sqrt{7}}{2} \leq \frac{5+3}{2} = 4 \end{align*}
• Since the volume is zero when \(x=0,3\text{,}\) it must be the case that the volume is maximised when \(x = x_- = \frac{5 - \sqrt{7}}{2}\text{.}\)
• Notice that since \(0 \lt x_- \lt 3\) we know that the other lengths are positive, so our answer makes sense. Further, the question only asks for the length \(x\) and not the resulting volume so we have answered the question.

Find the point on the line \(y=6-3x\) that is closest to the point \((7,5)\text{.}\)

Solution In this problem

• A simple picture

  
• Some notation is already given to us. Let a point on the line have coordinates \((x,y)\text{,}\) and we do not need units. And let \(\ell\) be the distance from the point \((x,y)\) to the point \((7,5)\text{.}\)
• Since the points are on the line the coordinates \((x,y)\) must obey
  \begin{gather*} y=6-3x \end{gather*}
  Notice that \(x\) and \(y\) have no further constraints. The distance \(\ell\) is given by
  \begin{align*} \ell^2 &= (x-7)^2 + (y-5)^2 \end{align*}
• We can now eliminate the variable \(y\text{:}\)
  \begin{align*} \ell^2 &= (x-7)^2 + (y-5)^2\\ &= (x-7)^2 + (6-3x-5)^2 = (x-7)^2 + (1-3x)^2\\ &= x^2-14x+49 + 1-6x+9x^2 = 10x^2-20x+50\\ &= 10(x^2-2x+5)\\ \ell &= \sqrt{10} \cdot \sqrt{x^2-2x+5} \end{align*}
  Notice that as \(x \to \pm \infty\) the distance \(\ell \to +\infty\text{.}\)

• We can now apply Theorem 3.5.17

  • Since the distance is defined for all real \(x\text{,}\) we do not have to check the endpoints of the domain — there are none.
  • Form the derivative:
    \begin{align*} \diff{\ell}{x} &= \sqrt{10} \frac{2x-2}{2\sqrt{x^2-2x+5}} \end{align*}
    It is zero when \(x=1\text{,}\) and undefined if \(x^2-2x+5 \lt 0\text{.}\) However, since
    \begin{align*} x^2-2x+5 &= (x^2-2x+1)+4 = \underbrace{(x-1)^2}_{\geq0}+4 \end{align*}
    we know that \(x^2-2x+5 \geq 4\text{.}\) Thus the function has no singular points and the only critical point occurs at \(x=1\text{.}\) The corresponding function value is then
    \begin{align*} \ell(1) &=\sqrt{10} \sqrt{1-2+5} = 2 \sqrt{10}. \end{align*}
  • Thus the minimum value of the distance is \(\ell=2\sqrt{10}\) and occurs at \(x=1\text{.}\)
• This answer makes sense — the distance is not negative.

• The question asks for the point that minimises the distance, not that minimum distance. Hence the answer is \(x=1, y=6-3 = 3\text{.}\) I.e.

  The point that minimises the distance is \((1,3)\text{.}\)

Notice that we can make the analysis easier by observing that the point that minimises the distance also minimises the squared-distance. So that instead of minimising the function \(\ell\text{,}\) we can just minimise \(\ell^2\text{:}\)

The resulting algebra is a bit easier and we don't have to hunt for singular points.

Find the minimum distance from \((2,0)\) to the curve \(y^2=x^2+1\text{.}\)

Solution This is very much like the previous question.

• After reading the problem carefully we can draw a picture

  
• In this problem we do not need units and the variables \(x,y\) are supplied. We define the distance to be \(\ell\) and it is given by
  \begin{align*} \ell^2 &= (x-2)^2+y^2. \end{align*}
  As noted in the previous problem, we will minimise the squared-distance since that also minimises the distance.
• Since \(x,y\) satisfy \(y^2=x^2+1\text{,}\) we can write the distance as a function of \(x\text{:}\)
  \begin{align*} \ell^2 &= (x-2)^2 + y^2 = (x-2)^2 + (x^2+1) \end{align*}
  Notice that as \(x \to \pm \infty\) the squared-distance \(\ell^2 \to +\infty\text{.}\)
• Since the squared-distance is a polynomial it will not have any singular points, only critical points. The derivative is
  \begin{align*} \diff{}{x} \ell^2 &= 2(x-2) + 2x = 4x-4 \end{align*}
  so the only critical point occurs at \(x=1\text{.}\)
• When \(x=1, y=\pm \sqrt{2}\) and the distance is
  \begin{align*} \ell^2 &= (1-2)^2 + (1+1) = 3 & \ell=\sqrt{3} \end{align*}
  and thus the minimum distance from the curve to \((2,0)\) is \(\sqrt{3}\text{.}\)

A water trough is to be constructed from a metal sheet of width \(45\) cm by bending up one third of the sheet on each side through an angle \(\theta\text{.}\) Which \(\theta\) will allow the trough to carry the maximum amount of water?

Solution Clearly \(0 \leq \theta \leq \pi\text{,}\) so we are back in the domain ⁸Again, no pun intended. of Corollary 3.5.13.

• After reading the problem carefully we should realise that it is really asking us to maximise the cross-sectional area. A figure really helps.

  
• From this we are led to define the height \(h\) \(cm\) and cross-sectional area \(A\) \(cm^2\text{.}\) Both are functions of \(\theta\text{.}\)
  \begin{align*} h &= 15 \sin \theta \end{align*}
  while the area can be computed as the sum of the central \(15 \times h\) rectangle, plus two triangles. Each triangle has height \(h\) and base \(15 \cos \theta\text{.}\) Hence
  \begin{align*} A &= 15h + 2 \cdot \frac{1}{2} \cdot h \cdot 15 \cos \theta\\ &= 15h \left(1 + \cos \theta \right) \end{align*}
• Since \(h = 15\sin \theta\) we can rewrite the area as a function of just \(\theta\text{:}\)
  \begin{align*} A(\theta) &= 225 \sin\theta \left(1 + \cos \theta \right) \end{align*}
  where \(0 \leq \theta \leq \pi\text{.}\)
• Now we use Corollary 3.5.13. The ends of the interval give
  \begin{align*} A(0) &= 225 \sin 0 (1 + \cos 0) = 0\\ A(\pi) &= 225 \sin \pi ( 1+ \cos \pi) = 0 \end{align*}
  The derivative is
  \begin{align*} A'(\theta) &= 225 \cos\theta \cdot (1+\cos\theta) + 225 \sin\theta \cdot(-\sin\theta)\\ &= 225 \left[ \cos \theta + \cos^2\theta - \sin^2\theta \right] \qquad\text {recall }\sin^2\theta = 1 - \cos^2\theta\\ &= 225 \left[ \cos\theta +2\cos^2\theta -1 \right] \end{align*}
  This is a continuous function, so there are no singular points. However we can still hunt for critical points by solving \(A'(\theta) = 0\text{.}\) That is
  \begin{align*} 2\cos^2\theta + \cos \theta -1 &= 0 & \text{factor carefully}\\ (2\cos\theta -1)(\cos \theta+1) &= 0 \end{align*}
  Hence we must have \(\cos \theta =-1\) or \(\cos\theta = \half\text{.}\) On the domain \(0\leq \theta \leq \pi\text{,}\) this means \(\theta = \pi/3\) or \(\theta = \pi\text{.}\)
  \begin{align*} A(\pi) &= 0\\ A(\pi/3) &= 225 \sin(\pi/3) (1 + \cos(\pi/3))\\ & = 225 \cdot \frac{\sqrt{3}}{2} \cdot \left( 1 + \frac{1}{2} \right)\\ &= 225 \cdot \frac{3\sqrt{3}}{4} \approx 292.28 \end{align*}
• Thus the cross-sectional area is maximised when \(\theta = \dfrac{\pi}{3}\text{.}\)

Find the points on the ellipse \(\frac{x^2}{4}+y^2=1\) that are nearest to and farthest from the point \((1,0)\text{.}\)

Solution While this is another distance problem, the possible values of \(x,y\) are bounded, so we need Corollary 3.5.13 rather than Theorem 3.5.17.

• We start by drawing a picture:

  
• Let \(\ell\) be the distance from the point \((x,y)\) on the ellipse to the point \((1,0)\text{.}\) As was the case above, we will maximise the squared-distance.
  \begin{align*} \ell^2 &= (x-1)^2 + y^2. \end{align*}

• Since \((x,y)\) lie on the ellipse we have

  \begin{gather*} \frac{x^2}{4}+y^2=1 \end{gather*}

  Note that this also shows that \(-2 \leq x \leq 2\) and \(-1 \leq y \leq 1\text{.}\)

  Isolating \(y^2\) and substituting this into our expression for \(\ell^2\) gives

  \begin{align*} \ell^2 &= (x-1)^2 + \underbrace{1-x^2/4}_{=y^2}. \end{align*}
• Now we can apply Corollary 3.5.13. The endpoints of the domain give
  \begin{align*} \ell^2(-2) &= (-2-1)^2 + 1 - (-2)^2/4 = 3^2+1-1 = 9\\ \ell^2(2) &= (2-1)^2 + 1 - 2^2/4 = 1+1-1 = 1 \end{align*}
  The derivative is
  \begin{align*} \diff{}{x} \ell^2 &= 2(x-1) - x/2 = \frac{3x}{2} - 2 \end{align*}
  Thus there are no singular points, but there is a critical point at \(x = 4/3\text{.}\) The corresponding squared-distance is
  \begin{align*} \ell^2(4/3) &= \left( \frac{4}{3}-1\right)^2 +1 - \frac{(4/3)^2}{4}\\ &= (1/3)^2 + 1 - (4/9) = 6/9 = 2/3. \end{align*}

• To summarise (and giving distances and coordinates of points):

  \(x\)	\((x,y)\)	\(\ell\)
  \(-2\)	\((-2,0)\)	\(3\)
  \(\frac{4}{3}\)	\(\big(\frac{4}{3},\pm\frac{\sqrt{5}}{3}\big)\)	\(\sqrt{\frac{2}{3}}\)
  \(2\)	\((2,0)\)	\(1\)

  The point of maximum distance is \((-2,0)\text{,}\) and the point of minimum distance is \(\left(\frac{4}{3},\pm\frac{\sqrt{5}}{3}\right)\text{.}\)

Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side \(a\) if one side of the rectangle lies on the base of the triangle.

Solution Since the rectangle must sit inside the triangle, its dimensions are bounded and we will end up using Corollary 3.5.13.

• Carefully draw a picture:

  

  

  We have drawn (on the left) the triangle in the \(xy\)-plane with its base on the \(x\)-axis. The base has been drawn running from \((-a/2,0)\) to \((a/2,0)\) so its centre lies at the origin. A little Pythagoras (or a little trigonometry) tells us that the height of the triangle is

  \begin{align*} \sqrt{a^2-(a/2)^2} &= \frac{\sqrt{3}}{2}\cdot a = a\cdot \sin\frac{\pi}{3} \end{align*}

  Thus the vertex at the top of the triangle lies at \(\left(0,\frac{\sqrt{3}}{2}\cdot a\right)\text{.}\)

• If we construct a rectangle that does not touch the sides of the triangle, then we can increase the dimensions of the rectangle until it touches the triangle and so make its area larger. Thus we can assume that the two top corners of the rectangle touch the triangle as drawn in the right-hand figure above.
• Now let the rectangle be \(2x\) wide and \(y\) high. And let \(A\) denote its area. Clearly
  \begin{align*} A &= 2xy. \end{align*}
  where \(0 \leq x \leq a/2\) and \(0\leq y\leq \frac{\sqrt{3}}{2}a\text{.}\)
• Our construction means that the top-right corner of the rectangle will have coordinates \((x,y)\) and lie on the line joining the top vertex of the triangle at \((0,\sqrt{3}a/2)\) to the bottom-right vertex at \((a/2,0)\text{.}\) In order to write the area as a function of \(x\) alone, we need the equation for this line since it will tell us how to write \(y\) as a function of \(x\text{.}\) The line has slope
  \begin{align*} \text{slope} &= \frac{\sqrt{3}a/2 - 0}{0-a/2} = -\sqrt{3}. \end{align*}
  and passes through the point \((0,\sqrt{3}a/2)\text{,}\) so any point \((x,y)\) on that line satisfies:
  \begin{align*} y &= -\sqrt{3}x + \frac{\sqrt{3}}{2}a. \end{align*}
• We can now write the area as a function of \(x\) alone
  \begin{align*} A(x) &= 2x \left( -\sqrt{3}x + \frac{\sqrt{3}}{2}a \right)\\ &= \sqrt{3} x(a-2x). \end{align*}
  with \(0\leq x \leq a/2\text{.}\)
• The ends of the domain give:
  \begin{align*} A(0) &= 0 & A(a/2) &= 0. \end{align*}
  The derivative is
  \begin{align*} A'(x) &= \sqrt{3} \left( x \cdot (-2) + 1 \cdot (a-2x) \right) = \sqrt{3} (a-4x). \end{align*}
  Since this is a polynomial there are no singular points, but there is a critical point at \(x=a/4\text{.}\) There
  \begin{align*} A(a/4) &= \sqrt{3} \cdot \frac{a}{4} \cdot (a - a/2) = \sqrt{3} \cdot \frac{a^2}{8}.\\ y &= -\sqrt{3}\cdot (a/4) + \frac{\sqrt{3}}{2} a = \sqrt{3}\cdot \frac{a}{4}. \end{align*}

• Checking the question again, we see that we are asked for the dimensions rather than the area, so the answer is \(2x \times y\text{:}\)

  The largest such rectangle has dimensions \(\frac{a}{2} \times \frac{\sqrt{3} a}{4}\text{.}\)

Consider the figure below which shows the trajectory of a ray of light as it passes through two different mediums (say air and water).

Let \(c_a\) be the speed of light in air and \(c_w\) be the speed of light in water. Fermat's principle states that a ray of light will always travel along a path that minimises the time taken. So if a ray of light travels from \(P\) (in air) to \(Q\) (in water) then it will “choose” the point \(O\) (on the interface) so as to minimise the total time taken. Use this idea to show Snell's law,

where \(\theta_i\) is the angle of incidence and \(\theta_r\) is the angle of refraction (as illustrated in the figure above).

Solution This problem is a little more abstract than the others we have examined, but we can still apply Theorem 3.5.17.

• We are given a figure in the statement of the problem and it contains all the relevant points and angles. However it will simplify things if we decide on a coordinate system. Let's assume that the point \(O\) lies on the \(x\)-axis, at coordinates \((x,0)\text{.}\) The point \(P\) then lies above the axis at \((X_P,+Y_P)\text{,}\) while \(Q\) lies below the axis at \((X_Q,-Y_Q)\text{.}\) This is drawn below.

  
• The statement of Snell's law contains terms \(\sin \theta_i\) and \(\sin \theta_r\text{,}\) so it is a good idea for us to see how to express these in terms of the coordinates we have just introduced:
  \begin{align*} \sin \theta_i &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{(x-X_P)}{\sqrt{(X_P-x)^2 + Y_P^2}}\\ \sin \theta_r &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{(X_Q-x)}{\sqrt{(X_Q-x)^2 + Y_Q^2}} \end{align*}
• Let \(\ell_P\) denote the distance \(PO\text{,}\) and \(\ell_Q\) denote the distance \(OQ\text{.}\) Then we have
  \begin{align*} \ell_P &= \sqrt{(X_P-x)^2+Y_P^2}\\ \ell_Q &= \sqrt{(X_Q-x)^2+Y_Q^2} \end{align*}
  If we then denote the total time taken by \(T\text{,}\) then
  \begin{align*} T &= \frac{\ell_P}{c_a} + \frac{\ell_Q}{c_w} = \frac{1}{c_a}\sqrt{(X_P-x)^2+Y_P^2} + \frac{1}{c_w}\sqrt{(X_Q-x)^2+Y_Q^2} \end{align*}
  which is written as a function of \(x\) since all the other terms are constants.
• Notice that as \(x \to +\infty\) or \(x\to-\infty\) the total time \(T \to \infty\) and so we can apply Theorem 3.5.17. The derivative is
  \begin{align*} \diff{T}{x} &= \frac{1}{c_a} \frac{-2(X_P-x)}{2\sqrt{(X_P-x)^2+Y_P^2}} + \frac{1}{c_w} \frac{-2(X_Q-x)}{2\sqrt{(X_Q-x)^2+Y_Q^2}} \end{align*}
  Notice that the terms inside the square-roots cannot be zero or negative since they are both sums of squares and \(Y_P,Y_Q \gt 0\text{.}\) So there are no singular points, but there is a critical point when \(T'(x) = 0\text{,}\) namely when
  \begin{align*} 0 &= \frac{1}{c_a} \frac{X_P-x}{\sqrt{(X_P-x)^2+Y_P^2}} + \frac{1}{c_w} \frac{X_Q-x}{\sqrt{(X_Q-x)^2+Y_Q^2}}\\ &= \frac{-\sin\theta_i}{c_a} + \frac{\sin\theta_r}{c_w} \end{align*}
  Rearrange this to get
  \begin{align*} \frac{\sin\theta_i}{c_a} &= \frac{\sin\theta_r}{c_w}\\ \end{align*}

  move sines to one side

  \begin{align*} \frac{\sin\theta_i}{\sin \theta_r} &= \frac{c_a}{c_w} \end{align*}
  which is exactly Snell's law.

The Statue of Liberty has height \(46\)m and stands on a \(47\)m tall pedestal. How far from the statue should an observer stand to maximize the angle subtended by the statue at the observer's eye, which is \(1.5\)m above the base of the pedestal?

Solution Obviously if we stand too close then all the observer sees is the pedestal, while if they stand too far then everything is tiny. The best spot for taking a photograph is somewhere in between.

• Draw a careful picture ¹¹And make some healthy use of public domain clip art.

  

  and we can put in the relevant lengths and angles.

• The height of the statue is \(h = 46\)m, and the height of the pedestal (above the eye) is \(p = 47-1.5 = 45.5\)m. The horizontal distance from the statue to the eye is \(x\text{.}\) There are two relevant angles. First \(\theta\) is the angle subtended by the statue, while \(\varphi\) is the angle subtended by the portion of the pedestal above the eye.
• Some trigonometry gives us
  \begin{align*} \tan \varphi &= \frac{p}{x}\\ \tan (\varphi+\theta) &= \frac{p+h}{x} \end{align*}
  Thus
  \begin{align*} \varphi &= \arctan \frac{p}{x}\\ \varphi+\theta &= \arctan \frac{p+h}{x} \end{align*}
  and so
  \begin{align*} \theta &= \arctan \frac{p+h}{x} - \arctan\frac{p}{x}. \end{align*}

• If we allow the viewer to stand at any point in front of the statue, then \(0\le x \lt \infty\text{.}\) Further observe that as \(x \rightarrow \infty\) or \(x \rightarrow 0\) the angle \(\theta \rightarrow 0\text{,}\) since

  \begin{equation*} \lim_{x\rightarrow\infty} \arctan \frac{p+h}{x} = \lim_{x\rightarrow\infty} \arctan \frac{p}{x} = 0 \end{equation*}

  and

  \begin{equation*} \lim_{x\rightarrow 0^+} \arctan \frac{p+h}{x} = \lim_{x\rightarrow 0^+} \arctan \frac{p}{x} = \frac{\pi}{2} \end{equation*}

  Clearly the largest value of \(\theta\) will be strictly positive and so has to be taken for some \(0 \lt x \lt \infty\text{.}\) (Note the strict inequalities.) This \(x\) will be a local maximum as well as a global maximum. As \(\theta\) is not singular at any \(0 \lt x \lt \infty\text{,}\) we need only search for critical points.

  A careful application of the chain rule shows that the derivative is

  \begin{align*} \diff{\theta}{x} &= \frac{1}{1+(\frac{p+h}{x})^2}\cdot \left(\frac{-(p+h)}{x^2}\right) - \frac{1}{1+(\frac{p}{x})^2}\cdot \left(\frac{-p}{x^2}\right)\\ &= \frac{-(p+h)}{x^2+(p+h)^2} + \frac{p}{x^2+p^2} \end{align*}

  So a critical point occurs when

  \begin{align*} \frac{(p+h)}{x^2+(p+h)^2} &= \frac{p}{x^2+p^2} & \text{cross multiply}\\ (p+h)(x^2+p^2) &= p(x^2+(p+h)^2) & \text{collect $x$ terms}\\ x^2 (p+h-p) &= p(p+h)^2 - p^2 (p+h) & \text{clean up}\\ h x^2 &= p(p+h)(p+h-p) = ph(p+h) \\ \amp\hskip1in \text{cancel common factors}\\ x^2 &= p(p+h)\\ x &= \pm\sqrt{p(p+h)} \approx \pm 64.9m \end{align*}
• Thus the best place to stand approximately \(64.9\)m in front or behind the statue. At that point \(\theta \approx 0.348\) radians or \(19.9^\circ\text{.}\)

Find the length of the longest rod that can be carried horizontally (no tilting allowed) from a corridor \(3\)m wide into a corridor \(2\)m wide. The two corridors are perpendicular to each other.

Solution

• Suppose that we are carrying the rod around the corner, then if the rod is as long as possible it must touch the corner and the outside walls of both corridors. A picture of this is show below.

  

  You can see that this gives rise to two similar triangles, one inside each corridor. Also the maximum length of the rod changes with the angle it makes with the walls of the corridor.

• Suppose that the angle between the rod and the inner wall of the \(3\)m corridor is \(\theta\text{,}\) as illustrated in the figure above. At the same time it will make an angle of \(\frac{\pi}{2}-\theta\) with the outer wall of the 2m corridor. Denote by \(\ell_1(\theta)\) the length of the part of the rod forming the hypotenuse of the upper triangle in the figure above. Similarly, denote by \(\ell_2(\theta)\) the length of the part of the rod forming the hypotenuse of the lower triangle in the figure above. Then
  \begin{gather*} \ell_1(\theta) = \frac{3}{\sin\theta}\qquad \ell_2(\theta) = \frac{2}{\cos\theta} \end{gather*}
  and the total length is
  \begin{gather*} \ell(\theta) = \ell_1(\theta) +\ell_2(\theta) = \frac{3}{\sin\theta}+\frac{2}{\cos\theta} \end{gather*}
  where \(0 \leq \theta \leq \frac{\pi}{2}\text{.}\)

• The length of the longest rod we can move through the corridor in this way is the minimum of \(\ell(\theta)\text{.}\) Notice that \(\ell(\theta)\) is not defined at \(\theta = 0, \frac{\pi}{2}\text{.}\) Indeed we find that as \(\theta \rightarrow 0^+\) or \(\theta\rightarrow \frac{\pi}{2}^-\text{,}\) the length \(\ell\rightarrow +\infty\text{.}\) (You should be able to picture what happens to our rod in those two limits). Clearly the minimum allowed \(\ell(\theta)\) is going to be finite and will be achieved for some \(0 \lt \theta \lt \frac{\pi}{2}\) (note the strict inequalities) and so will be a local minimum as well as a global minimum. So we only need to find zeroes of \(\ell'(\theta)\text{.}\)

  Differentiating \(\ell\) gives

  \begin{align*} \diff{\ell}{\theta} &= -\frac{3 \cos\theta}{\sin^2\theta} + \frac{2\sin\theta}{\cos^2\theta} = \frac{-3\cos^3\theta +2 \sin^3\theta}{\sin^2\theta \cos^2\theta}. \end{align*}

  This does not exist at \(\theta = 0, \frac{\pi}{2}\) (which we have already analysed) but does exist at every \(0 \lt \theta \lt \frac{\pi}{2}\) and is equal to zero when the numerator is zero. Namely when

  \begin{align*} 2\sin^3\theta &= 3 \cos^3 \theta & \text{divide by $\cos^3 \theta$}\\ 2 \tan^3\theta &= 3\\ \tan\theta &= \sqrt[3]{\frac{3}{2}} \end{align*}

• From this we can recover \(sin\theta\) and \(cos\theta\text{,}\) without having to compute \(\theta\) itself. We can, for example, construct a right-angle triangle with adjacent length \(\sqrt[3]{2}\) and opposite length \(\sqrt[3]{3}\) (so that \(\tan\theta=\sqrt[3]{3/2}\)):

  

  It has hypotenuse \(\sqrt{ 3^{2/3} + 2^{2/3}}\text{,}\) and so

  \begin{align*} \sin \theta &= \frac{3^{1/3}}{\sqrt{3^{2/3}+2^{2/3}}}\\ \cos \theta &= \frac{2^{1/3}}{\sqrt{3^{2/3}+2^{2/3}}} \end{align*}

  Alternatively could use the identities:

  \begin{align*} 1 + \tan^2\theta &= \sec^2\theta & 1+\cot^2\theta &= \csc^2\theta \end{align*}

  to obtain expressions for \(1/\cos\theta\) and \(1/\sin\theta\text{.}\)

• Using the above expressions for \(\sin\theta, \cos\theta\) we find the minimum of \(\ell\) (which is the longest rod that we can move):
  \begin{align*} \ell &= \frac{3}{\sin\theta} + \frac{2}{\cos\theta} =\frac{3}{ \frac{\root 3\of 3} {\sqrt{2^{\frac{2}{3}}+3^{\frac{2}{3}}}} } +\frac{2}{ \frac{\root 3\of 2}{\sqrt{2^{\frac{2}{3}}+3^{\frac{2}{3}}}} }\\ & =\sqrt{2^{\frac{2}{3}}+3^{\frac{2}{3}}} \big[3^{\frac{2}{3}}+2^{\frac{2}{3}}\big]\\ & ={\big[2^{\frac{2}{3}}+3^{\frac{2}{3}}\big]}^{\frac{3}{2}} \approx 7.02\text{m} \end{align*}