One important application of differential calculus is to find the maximum (or minimum) value of a function. This often finds real world applications in problems such as the following.
Solution We will describe a general approach to these sorts of problems in Sections 3.5.2 and 3.5.3 below, but here we can take a stab at starting the problem.
Begin by defining variables and their units (more generally we might draw a picture too); let the dimensions of the paddock be \(x\) by \(y\) metres.
At this stage we cannot apply the calculus we have developed since the area is a function of two variables and we only know how to work with functions of a single variable. We need to eliminate one variable.
We know that the perimeter of the rectangle (and hence the dimensions \(x\) and \(y\)) are constrained by the amount of fencing materials the farmer has to hand:
\begin{align*}
2x+2y &\leq 400\\
\end{align*}
and so we have
\begin{align*}
y &\leq 200-x
\end{align*}
Clearly the area of the paddock is maximised when we use all the fencing possible, so
The above example is sufficiently simple that we can likely determine the answer by several different methods. In general, we will need more systematic methods for solving problems of the form
Notice that we have not yet made the ideas of maximum and minimum very precise. For the moment think of maximum as “the biggest value” and minimum as “the smallest value”.
It is important to distinguish between “the smallest value” and “the smallest magnitude”. For example, because
\begin{gather*}
-5 \lt -1
\end{gather*}
the number \(-5\) is smaller than \(-1\text{.}\) But the magnitude of \(-1\text{,}\) which is \(|-1|=1\text{,}\) is smaller than the magnitude of \(-5\text{,}\) which is \(|-5|=5\text{.}\) Thus the smallest number in the set \(\{-1, -5\}\) is \(-5\text{,}\) while the number in the set \(\{-1,-5\}\) that has the smallest magnitude is \(-1\text{.}\)
Now back to thinking about what happens around a maximum. Suppose that the maximum value of \(f(x)\) is \(f(c)\text{,}\) then for all “nearby” points, the function should be smaller.
Firstly, in order for the argument to work we only need that \(f(x) \lt f(c)\) for \(x\) close to \(c\) — it does not matter what happens for \(x\) values far from \(c\text{.}\)
Secondly, in the above argument we needed to consider \(f(x)\) for \(x\) both to the left of and to the right of \(c\text{.}\) If the function \(f(x)\) is defined on a closed interval \([a,b]\text{,}\) then the above argument only applies when \(a \lt c \lt b\) — not when \(c\) is either of the endpoints \(a\) and \(b\text{.}\)
This function has only 1 maximum value (the middle green point in the graph) and 1 minimum value (the rightmost blue point), however it has 4 points at which the derivative is zero. In the small intervals around those points where the derivative is zero, we can see that function is locally a maximum or minimum, even if it is not the global maximum or minimum. We clearly need to be more careful distinguishing between these cases.
Let \(I\) be an interval, like \((a,b)\) or \([a,b]\) for example, and let the function \(f(x)\) be defined for all \(x \in I\text{.}\) Now let \(c\in I\text{.}\) Then
we say that \(f(x)\) has a global (or absolute) minimum on the interval \(I\) at the point \(x=c\) if \(f(x)\ge f(c)\) for all \(x\in I\text{.}\)
Beware that, while many textbooks use these definitions of local minimum and maximum, some textbooks exclude the endpoints \(a\text{,}\)\(b\) of the interval \([a,b]\) from their definitions. Our definitions allow the endpoints \(a\) and \(b\) to be local minima and maxima. Note that, under our definitions, every global minimum (maximum) is also a local minimum (maximum).
minimum on \(I\) at \(x=c\) if \(f(x)\ge f(c)\) for all \(x\in I\) that are near \(c\text{.}\) Precisely, if there is a \(\delta>0\) such that \(f(x)\ge f(c)\) for all \(x\in I\) that are within a distance \(\delta\) of \(c\text{.}\)
Similarly, we say that \(f(x)\) has a local maximum on \(I\) at \(x=c\) if \(f(x)\le f(c)\) for all \(x\in I\) that are near \(c\text{.}\) Precisely, if there is a \(\delta>0\) such that \(f(x)\le f(c)\) for all \(x\in I\) that are within a distance \(\delta\) of \(c\text{.}\)
The global maxima and minima of a function are called the global extrema of the function, while the local maxima and minima are called the local extrema.
It has 3 local maxima and 3 local minima on the interval \([a,b]\text{.}\) The global maximum occurs at the middle green point (which is also a local maximum), and the global minimum occurs at the rightmost blue point (which is also a local minimum).
Using the above definition we can summarise what we have learned above as the following theorem 2
This is one of several important mathematical contributions made by Pierre de Fermat, a French government lawyer and amateur mathematician, who lived in the first half of the seventeenth century.
Let the function \(f(x)\) be defined on the interval \(I\) and let \(a\text{,}\)\(b\text{,}\)\(c\) be points in \(I\) with \(a\lt c\lt b\text{.}\) If \(f(x)\) has a local maximum or local minimum at \(x=c\) and if \(f'(c)\) exists, then \(f'(c)=0\text{.}\)
It is often (but not always) the case that, when \(f(x)\) has a local maximum at \(x=c\text{,}\) the function \(f(x)\) increases strictly as \(x\) approaches \(c\) from the left and decreases strictly as \(x\) leaves \(c\) to the right. That is, \(f'(x) \gt 0\) for \(x\) just to the left of \(c\) and \(f'(x) \lt 0\) for \(x\) just to the right of \(c\text{.}\) Then, it is often the case, because \(f'(x)\) is decreasing as \(x\) increases through \(c\text{,}\) that \(f''(c) \lt 0\text{.}\)
Conversely, if \(f'(c)=0\) and \(f''(c) \lt 0\text{,}\) then, just to the right of \(c\text{,}\)\(f'(x)\) must be negative, so that \(f(x)\) is decreasing, and just to the left of \(c\text{,}\)\(f'(x)\) must be positive, so that \(f(x)\) is increasing. So \(f(x)\) has a local maximum at \(c\text{.}\)
Similarly, it is often the case that, when \(f(x)\) has a local minimum at \(x=c\text{,}\)\(f'(x) \lt 0\) for \(x\) just to the left of \(c\) and \(f'(x) \gt 0\) for \(x\) just to the right of \(c\) and \(f''(x) \gt 0\text{.}\)
Conversely, if \(f'(c)=0\) and \(f''(c) \gt 0\text{,}\) then, just to the right of \(c\text{,}\)\(f'(x)\) must be positive, so that \(f(x)\) is increasing, and, just to the left of \(c\text{,}\)\(f'(x)\) must be negative, so that \(f(x)\) is decreasing. So \(f(x)\) has a local minimum at \(c\text{.}\)
Observe that, in this example, \(f'(x)\) changes continuously from negative to positive at the local minimum, taking the value zero at the local minimum (the red dot).
Observe that, in this example, \(f'(x)\) changes discontinuously from negative to positive at the local minimum (\(x=0\)) and \(f'(0)\) does not exist.
The point \(c\) is an endpoint of the interval \(I\text{.}\) This case is also illustrated in the above figure. The endpoints \(a\) and \(b\) are both local maxima. But \(f'(a)\) and \(f'(b)\) are not zero.
This theorem demonstrates that the points at which the derivative is zero or does not exist are very important. It simplifies the discussion that follows if we give these points names.
if \(f'(c)\) does not exist then we call \(x=c\) a singular point 3
For \(c\) to be a local maximum or minimum of \(f\text{,}\) the function \(f\) must obviously be defined at \(c\text{.}\) So here we are considering only points \(c\) in the domain of \(f\text{.}\) We will later, in §3.6.2, extend the definition of singular points of \(f\) to points that are not in the domain of \(f\text{.}\)
Note that some people (and texts) will combine both of these cases and call \(x=c\) a critical point when either the derivative is zero or does not exist. The reader should be aware of the lack of convention on this point 4
No pun intended.
and should be careful to understand whether the more inclusive definition of critical point is being used, or if the text is using the more precise definition that distinguishes critical and singular points.
We’ll now look at a few simple examples involving local maxima and minima, critical points and singular points. Then we will move on to global maxima and minima.
This derivative takes the value \(0\) at two different values of \(x\text{.}\) Namely \(x=c_-=-\sqrt{2}\) and \(x=c_+=\sqrt{2}\text{.}\) Here is a sketch of the graph of \(f(x)\text{.}\)
the global minimum of \(f(x)\text{,}\) for \(x\) in the interval \(-2\le x\le 3\text{,}\) is at \(x=c_+\) (i.e. we have \(f(x)\ge f(c_+)\) whenever \(-2\le x\le 3\)) and
the global maximum of \(f(x)\text{,}\) for \(x\) in the interval \(-2\le x\le 3\text{,}\) is at \(x=3\) (i.e. we have \(f(x)\le f(3)\) whenever \(-2\le x\le 3\)).
Note that we have carefully constructed this example to illustrate that the global maximum (or minimum) of a function on an interval may or may not also be a critical point of the function.
The graph of \(f(x)\) is sketched below. From that sketch we see that \(f(x)\) has neither a local maximum nor a local minimum at \(x=c\) despite the fact that \(f'(c)=0\) — we have \(f(x) \lt f(c)=0\) for all \(x \lt c=0\) and \(f(x) \gt f(c)=0\) for all \(x \gt c=0\text{.}\)
Note that this example has been constructed to illustrate that a critical point (or singular point) of a function need not be a local maximum or minimum for the function.
A very common error of logic that people make is “Affirming the consequent”. When the statement “if P then Q” is true, observing Q does not imply P. (“Affirming the consequent” eliminates “not” from the previous sentence.) For example, “If he is Shakespeare then he is dead.” and “That man is dead.” does not imply “He must be Shakespeare.”. Or you may have also seen someone use this reasoning: “If a person is a genius before their time then they are misunderstood.” “I am misunderstood.” “So I must be a genius before my time.”.
“Let \(\cdots\text{.}\) If \(f(x)\) has a local maximum/minimum at \(x=c\) and if \(f'(c)\) exists, then \(f'(c)=0\)”. It does not say that “if \(f'(c)=0\) then \(f\) has a local maximum/minimum at \(x=c\)”.
These derivatives never take the value \(0\text{,}\) so the functions \(f(x)\) and \(g(x)\) do not have any critical points. However both derivatives do not exist at the point \(x=0\text{,}\) so that point is a singular point for both \(f(x)\) and \(g(x)\text{.}\)
From the figures we see that both \(f(x)\) and \(g(x)\) have a local (and in fact global) minimum at \(x=0\) despite the fact that \(x=0\) is not a critical point.
Reread Theorem 3.5.4 yet again. It says “Let \(\cdots\text{.}\) If \(f(x)\) has a local maximum or local minimum at \(x=c\)and if \(f\) is differentiable at \(x=c\), then \(f'(c)=0\)”. It says nothing about what happens at points where the derivative does not exist. Indeed that is why we have to consider both critical points and singular points when we look for maxima and minima.
We now have a technique for finding local maxima and minima — just look at endpoints of the interval of interest and for values of \(x\) for which either \(f'(x)=0\) or \(f'(x)\) does not exist. What about finding global maxima and minima? We’ll start by stating explicitly that, under appropriate hypotheses, global maxima and minima are guaranteed to exist.
Let the function \(f(x)\) be defined and continuous on the closed, finite interval 6
The hypotheses that \(f(x)\) be continuous and that the interval be finite and closed are all essential. We suggest that you find three functions \(f_1(x)\text{,}\)\(f_2(x)\) and \(f_3(x)\) with \(f_1\) defined but not continuous on \(0\le x\le 1\text{,}\)\(f_2\) defined and continuous on \(-\infty \lt x \lt \infty\text{,}\) and \(f_3\) defined and continuous on \(0 \lt x \lt 1\text{,}\) and with none of \(f_1\text{,}\)\(f_2\) and \(f_3\) attaining either a global maximum or a global minimum.
\(-\infty \lt a\le x\le b \lt \infty\text{.}\) Then \(f(x)\) attains a maximum and a minimum at least once. That is, there exist numbers \(a\le x_m, x_M\le b\) such that
\begin{gather*}
f(x_m)\le f(x)\le f(x_M)
\qquad\text{for all }a\le x\le b
\end{gather*}
Suppose that the maximum (or minimum) value of \(f(x)\text{,}\) for \(a\le x\le b\text{,}\) is \(f(c)\text{.}\) What does that tell us about \(c\text{?}\)
If \(c\) obeys \(a \lt c \lt b\) (note the strict inequalities), then \(f\) has a local maximum (or minimum) at \(x=c\) and Theorem 3.5.4 tells us that either \(f'(c)=0\) or \(f'(c)\) does not exist. The only other place that a maximum or minimum can occur are at the ends of the interval. We can summarise this as:
Evaluate \(f(c)\) for each \(c\) in that list. The largest (or smallest) of those values is the largest (or smallest) value of \(f(x)\) for \(a\le x\le b\text{.}\)
Let’s now demonstrate how to use this strategy. The function in this first example is not too simple — but it is a good example of a function that contains both a singular point and a critical point.
Solution We will apply the method in Corollary 3.5.13. It is perhaps easiest to find the values at the endpoints of the intervals and then move on to the values at any critical or singular points.
Before we get into things, notice that we can rewrite the function by factoring it:
Notice that the numerator and denominator are defined for all \(x\text{.}\) The only place the derivative is undefined is when the denominator is zero. Hence the only singular point is at \(x=0\text{.}\) The corresponding function value is
Thus on the interval \(-1\leq x \leq 1\) the global maximum of \(f\) is \(5\text{,}\) and is taken at \(x=1\text{,}\) while the global minimum value of \(f(x)\) is \(0\text{,}\) and is taken at \(x=0\text{.}\)
As noted at the beginning of this section, the problem of finding maxima and minima is a very important application of differential calculus in the real world. We now turn to a number of examples of this process. But to guide the reader we will describe a general procedure to follow for these problems.
Variables — assign variables to the quantities in the problem along with their units. It is typically a good idea to make sensible choices of variable names: \(A\) for area, \(h\) for height, \(t\) for time etc.
Relations — find relations between the variables. By now you should know the quantity we are interested in (the one we want to maximise or minimise) and we need to establish a relation between it and the other variables.
Reduce — the relation down to a function of one variable. In order to apply the calculus we know, we must have a function of a single variable. To do this we need to use all the information we have to eliminate variables. We should also work out the domain of the resulting function.
Maximise or minimise — we can now apply the methods of Corollary 3.5.13 to find the maximum or minimum of the quantity we need (as the problem dictates).
Example3.5.15.Constructing a container of maximal volume.
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.
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 \(V m^3\text{.}\)
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?
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.
There is a new wrinkle in the next two examples. Each involves finding the minimum value of a function \(f(x)\) with \(x\) running over all real numbers, rather than just over a finite interval as in Corollary 3.5.13. Both in Example 3.5.18 and in Example 3.5.19 the function \(f(x)\) tends to \(+\infty\) as \(x\) tends to either \(+\infty\) or \(-\infty\text{.}\) So the minimum value of \(f(x)\) will be achieved for some finite value of \(x\text{,}\) which will be a local minimum as well as a global minimum.
If \(\displaystyle \lim_{x\rightarrow+\infty} f(x)=+\infty\) and \(\displaystyle \lim_{x\rightarrow-\infty} f(x)=+\infty\) and if \(f(x)\) has a global minimum at \(x = c\text{,}\) then there are 2 possibilities. Either
If \(\displaystyle \lim_{x\rightarrow+\infty} f(x)=-\infty\) and \(\displaystyle \lim_{x\rightarrow-\infty} f(x)=-\infty\) and if \(f(x)\) has a global maximum at \(x = c\text{,}\) then there are 2 possibilities. Either
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 \((7,5)\text{.}\)
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
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 \(\ell^2\text{:}\)
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?
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
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
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 \(\theta = \pi\text{.}\)
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.
Example3.5.22.Largest rectangle inside a triangle.
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.
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.
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
This next one is a good physics example. In it we will derive Snell’s Law 9
Snell’s law is named after the Dutch astronomer Willebrord Snellius who derived it in around 1621, though it was first stated accurately in 984 by Ibn Sahl.
from Fermat’s principle 10
Named after Pierre de Fermat who described it in a letter in 1662. The beginnings of the idea, however, go back as far as Hero of Alexandria in around 60CE. Hero is credited with many inventions including the first vending machine, and a precursor of the steam engine called an aeolipile.
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,
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:
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
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 11
And make some healthy use of public domain clip art.
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.
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
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.
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.
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 3m 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
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 \(\ell'(\theta)\text{.}\)
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
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 \(\tan\theta=\sqrt[3]{3/2}\)):
Identify every critical point and every singular point of \(f(x)\) on the graph below. Which correspond to local extrema? Which correspond to global extrema over the interval shown?
Below are a number of curves, all of which have a singular point at \(x=2\text{.}\) For each, label whether \(x=2\) is a local maximum, a local minimum, or neither.
Suppose \(f(x)\) is the constant function \(f(x)=4\text{.}\) What are the critical points and singular points of \(f(x)\text{?}\) What are its local and global maxima and minima?
Consider the function \(h(x)=x^3-12x+4\text{.}\) What are the coordinates of the local maximum of \(h(x)\text{?}\) What are the coordinates of the local minimum of \(h(x)\text{?}\)
Consider the function \(h(x)=2x^3-24x+1\text{.}\) What are the coordinates of the local maximum of \(h(x)\text{?}\) What are the coordinates of the local minimum of \(h(x)\text{?}\)
You are in a dune buggy at a point \(P\) in the desert, 12 km due south of the nearest point \(A\) on a straight east-west road. You want to get to a town \(B\) on the road \(18\) km east of \(A\text{.}\) If your dune buggy can travel at an average speed of 15 km/hr through the desert and 30 km/hr along the road, towards what point \(Q\) on the road should you head to minimize your travel time from \(P\) to \(B\text{?}\)
A closed three dimensional box is to be constructed in such a way that its volume is 4500 cm\({}^3\text{.}\) It is also specified that the length of the base is 3 times the width of the base. Find the dimensions of the box that satisfies these conditions and has the minimum possible surface area. Justify your answer.
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 metre, while the material for the other five sides costs $1 per square metre. Find the dimensions of the container which has the largest possible volume if the total cost of materials is $72.
A rectangle is inscribed in a semicircle of radius \(R\) so that one side of the rectangle lies along a diameter of the semicircle. Find the largest possible perimeter of such a rectangle, if it exists, or explain why it does not. Do the same for the smallest possible perimeter.
Find the maximal possible volume of a cylinder with surface area \(A\text{.}\) 12
Food is often packaged in cylinders, and companies wouldn’t want to waste the metal they are made out of. So, you might expect the dimensions you find in this problem to describe a tin of, say, cat food. Read here about why this isn’t the case.
What is the largest possible area of a window, with perimeter \(P\text{,}\) in the shape of a rectangle with a semicircle on top (so the diameter of the semicircle equals the width of the rectangle)?
Consider an open-top rectangular baking pan with base dimensions \(x\) centimetres by \(y\) centimetres and height \(z\) centimetres that is made from \(A\) square centimetres of tin plate. Suppose \(y = px\) for some fixed constant \(p\text{.}\)
Find the dimensions of the baking pan with the maximum capacity (i.e., maximum volume). Prove that your answer yields the baking pan with maximum capacity. Your answer will depend on the value of \(p\text{.}\)
Find the value of the constant \(p\) that yields the baking pan with maximum capacity and give the dimensions of the resulting baking pan. Prove that your answer yields the baking pan with maximum capacity.
A length of wire is cut into two pieces, one of which is bent to form a circle, the other to form a square. How should the wire be cut if the area enclosed by the two curves is maximized? How should the wire be cut if the area enclosed by the two curves is minimized? Justify your answers.
