Curvature of Curves

Returning to the process of successive differentiation, it may be asked: Why does anybody want to differentiate twice over? We know that when the variable quantities are space and time, by differentiating twice over we get the acceleration of a moving body, and that in the geometrical interpretation, as applied to curves, $\dfrac{dy}{dx}$ means the slope of the curve. But what can $\dfrac{d^2 y}{dx^2}$ mean in this case? Clearly it means the rate (per unit of length $x$) at which the slope is changing—in brief, it is an indication of the manner in which the slope of the portion of curve considered varies, that is, whether the slope of the curve increases or decreases when $x$ increases, or, in other words, whether the curve curves up or down towards the right.

Suppose a slope constant, as in the following figure.

Here, $\dfrac{dy}{dx}$ is of constant value.

Suppose, however, a case in which, like the next figure, the slope itself is getting greater upwards, then $\dfrac{d\left(\dfrac{dy}{dx}\right)}{dx}$, that is, $\dfrac{d^2y}{dx^2}$, will be positive.

If the slope is becoming less as you go to the right as in the following figure, then, even though the curve may be going upward, since the change is such as to diminish its slope, its $\dfrac{d^2y}{dx^2}$ will be negative.

It is now time to initiate you into another secret—how to tell whether the result that you get by “equating $\dfrac{dy}{dx}$ to zero” is a maximum or a minimum. The trick is this: After you have differentiated (so as to get the expression which you equate to zero), you then differentiate a second time, and look whether the result of the second differentiation is positive or negative. If $\dfrac{d^2y}{dx^2}$ comes out positive, then you know that the value of $y$ which you got was a minimum; but if $\dfrac{d^2y}{dx^2}$ comes out negative, then the value of $y$ which you got must be a maximum. That’s the rule.

The reason of it ought to be quite evident. Think of any curve that has a minimum point in it like the next figure, where the point of minimum $y$ is marked $M$, and the curve is concave upward.¹ To the left of $M$ the slope is downward, that is, negative, and is getting less negative. To the right of $M$ the slope has become upward, and is getting more and more upward. Clearly the change of slope as the curve passes through $M$ is such that $\dfrac{d^2y}{dx^2}$ is positive, for its operation, as $x$ increases toward the right, is to convert a downward slope into an upward one.

Similarly, consider any curve that has a maximum point in it (like Fig. 10.11 in this chapter), or like the next figure, where the curve is concave downward,² and the maximum point is marked $M$. In this case, as the curve passes through $M$ from left to right, its upward slope is converted into a downward or negative slope, so that in this case the “slope of the slope” $\dfrac{d^2y}{dx^2}$ is negative.

This is called the second derivative test for maxima and minima.

In summary:

If $\dfrac{d^2y}{dx^2}>0$, the curve is concave upward (or convex).

If $\dfrac{d^2y}{dx^2}<0$, the curve is concave downward (or concave).

and

The Second Derivative Test
Suppose $\dfrac{dy}{dx}=0$ for some particular value of $x$

If $\dfrac{d^2y}{dx^2}>0$ for this particular value of $x$, the value of $y$ for this $x$ is a minimum.

If $\dfrac{d^2y}{dx^2}<0$ for this particular value of $x$, the value of $y$ for this $x$ is a maximum.

Go back now to the examples of the last chapter and verify in this way the conclusions arrived at as to whether in any particular case there is a maximum or a minimum. You will find below a few worked out examples.

Example 12.1. Find the maximum or minimum of \[\text{(a)}\quad y = 4x^2-9x-6; \qquad \text{(b)}\quad y = 6 + 9x-4x^2;\] and ascertain if it be a maximum or a minimum in each case.

Solution. (a) \[\begin{align} \dfrac{dy}{dx}= 8x-9=0\quad \Rightarrow\quad x=1\frac{1}{8}. \end{align}\] The value of $y$ for (or at) $x=1\frac{1}{8}$ is $-11.065$. \[\dfrac{d^2y}{dx^2} = 8;\quad \text{it is~$+$; hence it is a minimum.}\] (b)

\[\dfrac{dy}{dx}= 9-8x=0\quad \Rightarrow\quad x = 1\frac{1}{8}.\] The value of $y$ for $x=1\frac{1}{8}$ is $-11.065$ \[\dfrac{d^2y}{dx^2}= -8;\quad \text{it is~$-$; hence it is a maximum.}\]

The graphs of $y=4x^2-9x-6$ and $y = 6 + 9x-4x^2$ are shown below.

Example 12.2. Find the maxima and minima of the function $y = x^3-3x+16$.

Solution. \[\dfrac{dy}{dx} = 3x^2 - 3 = 0\Rightarrow x^2 = 1\Rightarrow x = \pm1.\] \[\dfrac{d^2y}{dx^2} = 6x;\quad \text{for } x = 1; ~\dfrac{d^2y}{dx^2} \text{ is }+; \] hence $x=1$ corresponds to a minimum $y=14$. For $x=-1$, $\dfrac{d^2y}{dx^2}$ is $-$; hence $x=-1$ corresponds to a maximum $y=+18$.

The graph of $y = x^3-3x+16$ is shown below.

Example 12.3. Find the maxima and minima of $y=\dfrac{x-1}{x^2+2}$.

Solution. \[\frac{dy}{dx} = \frac{(x^2+2) \times 1 - (x-1) \times 2x}{(x^2+2)^2} = \frac{2x - x^2 + 2}{(x^2 + 2)^2} = 0;\] or $x^2 - 2x - 2 = 0$, whose solutions are $x =1+\sqrt{3}\approx+2.73$ and $x=1-\sqrt{3}\approx -0.73$. \[\begin{align} \dfrac{d^2y}{dx^2} &= - \frac{(x^2 + 2)^2 \times (2x-2) - (x^2 - 2x - 2)(4x^3 + 8x)}{(x^2 + 2)^4} \\ &= - \frac{2x^5 - 6x^4 - 8x^3 - 8x^2 - 24x + 8}{(x^2 + 2)^4}. \end{align}\]

The denominator is always positive, so it is sufficient to ascertain the sign of the numerator.

If we put $x = 2.73$, the numerator is negative; the maximum, $y = 0.183$.

If we put $x=-0.73$, the numerator is positive; the minimum, $y=-0.683$.

The graph of $y=\dfrac{x-1}{x^2+2}$ is shown below.

Example 12.4. The expense $C$ of handling the products of a certain factory varies with the weekly output $P$ according to the relation $C = aP + \dfrac{b}{c+P} + d$, where $a$, $b$, $c$, $d$ are positive constants. For what output will the expense be least?

Solution. \[\dfrac{dC}{dP} = a - \frac{b}{(c+P)^2} = 0\quad \text{for maximum or minimum;}\] hence $a = \dfrac{b}{(c+P)^2}$ and $P = \pm\sqrt{\dfrac{b}{a}} - c$.

As the output cannot be negative, $P=+\sqrt{\dfrac{b}{a}} - c$.

Now \[\frac{d^2C}{dP^2} = + \frac{b(2c + 2P)}{(c + P)^4},\] which is positive for all the values of $P$; hence $P = +\sqrt{\dfrac{b}{a}} - c$ corresponds to a minimum.

Example 12.5. The total cost per hour $C$ of lighting a building with $N$ lamps of a certain kind is \[C = N\left(\frac{C_l}{t} + \frac{EPC_e}{1000}\right),\] where $E$ is the commercial efficiency (watts per candle), \[\begin{align} &\text{$P$ is the candle power of each lamp,} \\ &\text{$t$ is the average life of each lamp in hours,} \\ &\text{$C_l =$ cost of renewal in cents per hour of use,} \\ &\text{$C_e =$ cost of energy per $1000$~watts per~hour.} \end{align}\]

Moreover, the relation connecting the average life of a lamp with the commercial efficiency at which it is run is approximately $t = mE^n$, where $m$ and $n$ are constants depending on the kind of lamp.

Find the commercial efficiency for which the total cost of lighting will be least.

Solution. We have \[\begin{align} C &= N\left(\frac{C_l}{m} E^{-n} + \frac{PC_e}{1000} E\right), \\ \dfrac{dC}{dE} &= \frac{PC_e}{1000} - \frac{nC_l}{m} E^{-(n+1)} = 0 \end{align}\] for maximum or minimum. \[E^{n+1} = \frac{1000 \times nC_l}{mPC_e},\] and \[E = \sqrt[ ^{\large n+1}]{\frac{1000 \times nC_l}{mPC_e}}\]

This is clearly for minimum, since \[\frac{d^2C}{dE^2} = (n + 1) \frac{nC_l}{m} E^{-(n+2)},\] which is positive for a positive value of $E$.

For a particular type of $16$ candle-power lamps, $C_l= 17$ cents, $C_e=5$ cents; and it was found that $m=10$ and $n=3.6$. \[E = \sqrt[^{\large 4.6}]{\frac{1000 \times 3.6 \times 17}{10 \times 16 \times 5}}=\left(\frac{1000 \times 3.6 \times 17}{10 \times 16 \times 5}\right)^{\frac{1}{4.6}} \approx 2.6\text{ watts per candle-power}.\]

Exercises

Exercise 12.1. Find the maxima and minima of \[y = x^3 + x^2 - 10x + 8.\]

Answer

Max.: $x \approx -2.19$, $y \approx 24.19$; min.:, $x \approx 1.52$, $y \approx -1.38$.

Solution

\[\begin{gathered} y=x^{3}+x^{2}-10 x+8 \\ \frac{d y}{d x}=3 x^{2}+2 x-10 \\ \frac{d y}{d x}=0 \Rightarrow x=\frac{-2 \pm \sqrt{4+120}}{6}=\frac{-2 \pm 2 \sqrt{31}}{6} \\ \text { Therefore } \frac{d y}{d x}=0 \text { if } x=-\frac{1+\sqrt{31}}{3} \approx-2.189 \text { or } x=\frac{\sqrt{31}-1}{3} \approx 1.522 \end{gathered}\]

To distinguish between a maximum and minimum, we find the second derivative:

\[\frac{d^{2} y}{d x^{2}}=6 x+2\]

When $x\approx -2.19$ \[\frac{d^{2} y}{d x^{2}}=-6 \times 2.19+2<0\] Therefore, the curve is concave downward close to $x\approx -2.19$ and $x\approx -2.19$ corresponds to a maximum $y \approx 24.19.$

When $x\approx1.52$ \[\begin{align} & y \approx-1.38 \\ & \frac{d^{2} y}{d x^{2}}=6 \times 1.52+2>0 \end{align}\] Therefore, the curve is concave upward near $x\approx1.52$ and $x\approx 1.52$ corresponds to a minimum $y \approx-1.38$.

This curve is shown below:

Exercise 12.2. Given $y = \dfrac{b}{a}x - cx^2$, find expressions for $\dfrac{dy}{dx}$, and for $\dfrac{d^2y}{dx^2}$, also find the value of $x$ which makes $y$ a maximum or a minimum, and show whether it is maximum or minimum (assume $c>0$).

Answer

$\dfrac{dy}{dx} = \dfrac{b}{a} - 2cx$; $\dfrac{d^2 y}{dx^2} = -2c$; $x = \dfrac{b}{2ac}$ (a maximum).

Solution

\[\begin{align} y & =\frac{b}{a} x-c x^{2} \\ \frac{d y}{d x} & =\frac{b}{a}-2 c x=0 \Rightarrow x=\frac{b}{2 a c} \\ \frac{d^{2} y}{d x^{2}} & =-2 c \end{align}\]

When $x=\frac{b}{2 a c}$

\[y=\frac{b}{a} \cdot \frac{b}{2 a c}-c \frac{b^{2}}{4 a^{2} c^{2}}=\frac{b^{2}}{4 a^{2} c}\] This is a maximum value since $\dfrac{d^{2} y}{d x^{2}}<0$.

Exercise 12.3. Find how many maxima and how many minima there are in the curve, the equation to which is \[y = 1 - \frac{x^2}{2} + \frac{x^4}{24};\] and how many in that of which the equation is \[y = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}.\]

Answer

(a) One maximum and two minima.
(b) One maximum. ($x = 0$; other points unreal.)

Solution

\[\begin{align} & y=1-\frac{x^{2}}{2}+\frac{x^{4}}{24} \\ & \frac{d y}{d x}=-x+\frac{1}{6} x^{3}=x\left(\frac{x^{2}}{6}-1\right) \\ & \frac{d y}{d x}=0 \quad \Leftrightarrow \quad x=0 \quad \text { or } \quad x= \pm \sqrt{6} \\ & \frac{d^{2} y}{d x^{2}}=-1+\frac{1}{2} x^{2} \end{align}\]

When $x=0$, \[\frac{d^{2} y}{d x^{2}}=-1<0\] Therefore, $x=0$ makes $y$ a maximum value. When $x=0$, $y=1$.

When $x=\sqrt{6}$ or $x=-\sqrt{6}$ \[\frac{d^{2} y}{d x^{2}}=-1+3=2>0\] Therefore, $x=\sqrt{6}\approx 2.449$ or $x=-\sqrt{6}\approx -2.449$ corresponds to a minimum $y=-\dfrac{1}{2}$.

The graph of $y=1 - \frac{x^2}{2} + \frac{x^4}{24}$ is shown below:

Now let’s consider the second function \[y =1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}.\] \[\begin{align} \frac{d y}{d x} & =-x+\frac{x^{3}}{6}-\frac{x^{5}}{120}=\frac{1}{120} x\left(-120+20 x^{2}-x^{4}\right) \\ \frac{d y}{d x}=0 & \Leftrightarrow x=0 \text { or } x^{4}-20 x^{2}+120=0 \end{align}\]

The expression $x^{4}-20 x^{2}+120$ is quadratic in terms of $t=x^{2}$

\[t^{2}-20 t+120=0\]

Since discriminant of this equation, $20^{2}-4 \times 120=-80$, is negative, the above equation has no roots.

Therefore,

\[\frac{d y}{d x}=0 \Leftrightarrow x=0\]

The value of $y$ for $x=0$ is 1 . Since

\[\frac{d^{2} y}{d x^{2}}=-1+\frac{x^{2}}{2}-\frac{x^{4}}{24}\]

is negative when $x=0$, $y=1$ is a maximum.

The graph of $y=1 - \frac{x^2}{2} + \frac{x^4}{24}-\frac{x^6}{720}$ is shown below:

Exercise 12.4. Find the maxima and minima of \[y=2x+1+\frac{5}{x^2}.\]

Answer

Min.: $x \approx 1.71$, $y \approx 6.13$.

Solution

\[y=2 x+1+\frac{5}{x^{2}}\] We can rewrite it as \[y=2 x+1+5 x^{-2}\] Then \[\frac{d y}{d x}=2-10 x^{-3}=2-\frac{10}{x^{3}}\] The second derivative is: \[\frac{d^{2} y}{d x^{2}}=30 x^{-4}=\frac{30}{x^{4}}\]

To find where $y$ is a maximum or minimum we set $\dfrac{dy}{dx}$ equal to zero: \[\frac{d y}{d x}=0 \Leftrightarrow x=\sqrt[3]{5}\]

Since $\dfrac{d^{2} y}{d x^{2}}>0$, $x=\sqrt[3]{5}$ corresponds to a minimum \[y=2 \sqrt[3]{5}+1+\frac{5}{5^ \frac{2}{3}} \approx 6.13.\]

Exercise 12.5. Find the maxima and minima of \[y=\frac{3}{x^2+x+1}.\]

Answer

Max: $x = -.5$, $y = 4$.

Solution

\[y=\frac{3}{x^{2}+x+1}\]

Using the Quotient Rule

\[\begin{align} & \frac{d y}{d x}=-\frac{3(2 x+1)}{\left(x^{2}+x+1\right)^{2}} \\ & \frac{d y}{d x}=0 \quad \Leftrightarrow \quad x=-\frac{1}{2} \end{align}\]

To determine if $y$ has a maximum or a minimum when $x=-\frac{1}{2}$, we can use the Second Derivative Test:

\[\frac{d^{2} y}{d x^{2}}=-3 \frac{2\left(x^{2}+x+1\right)^{2}-2(2 x+1)^{2}\left(x^{2}+x+1\right)}{\left(x^{2}+x+1\right)^{4}}.\]

When $x=-\frac{1}{2}$

\[\frac{d^{2} y}{d x^{2}}=-3 \frac{2 (+)-2\times 0 (\dots)}{(+)}=(-)\] Therefore, the curve is concave downward and $x=-\frac{1}{2}$ corresponds to a maximum $y=4$.

Alternatively, we can use the First Derivative Test.

When $x<\dfrac{1}{2}$, $\dfrac{d y}{d x}>0$, the curve ascends

When $x>\dfrac{1}{2}$, $\dfrac{d y}{d x}<0$, the curve descends.

Thus $y=4$ is a maximum that occurs when $x=-\frac{1}{2}$.

The graph of $y=\dfrac{3}{x^{2}+x+1}$ is shown below.

Exercise 12.6. Find the maxima and minima of \[y=\frac{5x}{2+x^2}.\]

Answer

Max.: $x \approx 1.414$, $y \approx 1.768$.
Min.: $x \approx -1.414$, $y \approx 1.768$.

Solution

\[y=\frac{5 x}{2+x^{2}}\]

Using the Quotient Rule:

\[\frac{d y}{d x}=\frac{5\left(2+x^{2}\right)-5 x(2 x)}{\left(2+x^{2}\right)^{2}}\]

\[\frac{d y}{d x}=\frac{10-5 x^{2}}{\left(2+x^{2}\right)^{2}}\]

\[\begin{align} & \frac{d y}{d x}=0 \Leftrightarrow 10-5 x^2=0 \\ & \frac{d y}{d x}=0 \Leftrightarrow \quad x= \pm \sqrt{2} \end{align}\]

Using the Second Derivative Test: \[\frac{d^2 y}{d x^2}=\frac{-10 x\left(2+x^2\right)^2-2(2 x)\left(2+x^2\right)\left(10-5 x^2\right)}{\left(2+x^2\right)^4}\]

When $x=\sqrt{2}$ \[\frac{d^2 y}{d r^2}=\frac{-10(+)(+)-2(+)(+)(0)}{(+)}=(-)\] It follows from the Second Derivative Test that $x=\sqrt{2}$ corresponds to a maximum $y=\dfrac{5\sqrt{2}}{2+2}\approx 1.768$.

When $x=-\sqrt{2}$ \[\frac{d^2 y}{d x^2}=\frac{-10(-)(+)-2(+)(+)(0)}{(+)}=(+)\] This means that $x=-\sqrt{2}$ corresponds to a minimum $y=-\dfrac{5\sqrt{2}}{4}\approx -1.768$.

The graph of $y=\dfrac{5 x}{2+x^{2}}$ is shown below.

Exercise 12.7. Find the maxima and minima of \[y=\frac{3x}{x^2-3} + \frac{x}{2} + 5.\]

Answer

Max.: $x \approx -3.565$, $y \approx 2.12$.
Min.: $x\approx +3.565$, $y \approx 7.88$.

Solution

\[\begin{align} & y=\frac{3 x}{x^{2}-3}+\frac{x}{2}+5 \\ \frac{d y}{d x} & =\frac{3\left(x^{2}-3\right)-6 x^{2}}{\left(x^{2}-3\right)^{2}}+\frac{1}{2} \\ & =\frac{-9-3 x^{2}}{\left(x^{2}-3\right)^{2}}+\frac{1}{2} \\ & =\frac{2\left(-9-3 x^{2}\right)+\left(x^{2}-3\right)^{2}}{2\left(x^{2}-3\right)^{2}} \\ & =\frac{-18-6 x^{2}+x^{4}-6 x^{2}+9}{2\left(x^{2}-3\right)^{2}} \\ & =\frac{x^{4}-12 x^{2}-9}{2\left(x^{2}-3\right)^{2}} \end{align}\] \[\frac{d y}{d x} =0 \quad\Leftrightarrow \quad x^4-12x^2-9=0.\] The equation $x^{4}-12 x^{2}-9=0$ is quadratic in terms of $x^{2}$. Thus \[x^{2}=\frac{12 \pm \sqrt{144+36}}{2}=\frac{12 \pm \sqrt{180}}{2}\] The negative is unacceptable because $x^{2} \geq 0$. Therefore

\[\begin{gathered} x^{2}=\frac{12+\sqrt{180}}{2}=3(\sqrt{5}+2) \\ \frac{d y}{d x}=0 \Leftrightarrow x=\sqrt{3(\sqrt{5}+2)} \approx 3.565\ \text{ or }\ x=-\sqrt{3(\sqrt{5}+2)} \approx-3.565 \end{gathered}\]

Using the Second Derivative Test \[\frac{d^{2} y}{d x^{2}}=\frac{2\left(3 x^{2}-24 x\right)\left(x^{2}-3\right)^{2}-8 x\left(x^{2}-3\right)\left(x^{4}-12 x^{2}-9\right)}{4\left(x^{2}-3\right)^{4}}\]

When $x=\sqrt{3(\sqrt{5}+2)} \approx 3.565$

\[\begin{align} \frac{d^{2} y}{d x^{2}} & =\frac{3\left(3 \times 3.565^{2}-24 \times 3.565\right)(+)-8(+)(-)(0)}{(+)} \\ & =\frac{3(-)(+)-0}{(+)} \\ & =(-) \end{align}\] Therefore, $x \approx 3.565$ corresponds to a maximum $y \approx 7.884$.

When $x \approx-3.565$

\[\frac{d^{2} y}{d x^{2}}=\frac{3(+)(+)-0}{(+)}=(+)\]

Therefore, $x \approx-3.565$ corresponds to a minimum $y \approx 2.116$.

The graph of $y=\dfrac{3x}{x^2-3} + \dfrac{x}{2} + 5$ is shown below.

Exercise 12.8. Divide a number $N$ into two parts in such a way that three times the square of one part plus twice the square of the other part shall be a minimum.

Answer

$0.4N$, $0.6N$.

Solution

Let

$x=$ part one

$z=$ part two

We know \[x+z=N\] and we want to minimize \[3 x^{2}+2 z^{2}\]

Since $z=N-x$, we want to minimize

\[3 x^{2}+2(N-x)^{2}\]

Let \[y =3 x^{2}+2(N-x)^{2}\] then \[\begin{align} \frac{d y}{d x} & =6 x+2 \times 2 \times(-1)(N-x) \\ & =10 x-4 N \end{align}\] \[\frac{d y}{d x}=0 \quad \Leftrightarrow \quad x=0.4 N\]

Does $x=0.4$ correspond to a minimum $y$ or a maximum $y$? To answer this, we can use the Second Derivative Test

\[\frac{d^{2} y}{d x^{2}}=10>0\]

It follows from the Second Derivative Test that $x=0.4 N$ corresponds to a minimum value.

When $x=0.4 N$, $z=0.6 N$ and

\[\begin{align} & y=3(0.4 N)^{2}+2(0.6 N)^{2} \\ & y=1.2 N^{2} \end{align}\]

Exercise 12.9. The efficiency $u$ of an electric generator at different values of output $x$ is expressed by the general equation: \[u=\frac{x}{a+bx+cx^2};\] where $a$ is a constant depending chiefly on the energy losses in the iron and $c$ a constant depending chiefly on the resistance of the copper parts. Find an expression for that value of the output at which the efficiency will be a maximum.

Answer

$x = \sqrt{\dfrac{a}{c}}$.

Solution

\[u=\frac{x}{a+b x+c x^{2}}\]

Using Quotient Rule:

\[\begin{align} \frac{d u}{d x} & =\frac{\left(a+b x+c x^{2}\right)-x(b+2 c x)}{\left(a+b x+c x^{2}\right)} \\ & =\frac{a-c x^{2}}{\left(a+b x+c x^{2}\right)^{2}} \end{align}\] \[\frac{d u}{d x} =0 \Leftrightarrow x= \pm \sqrt{\frac{c}{a}}\] \[\frac{d^{2} u}{d x^{2}} =\frac{-2 c x(a+b x+c x)^{2}+\left(a-c x^{2}\right)(b+2 c x)}{\left(a+b x+c x^{2}\right)^{4}}\]

When $x=\sqrt{\dfrac{c}{a}}$, \[\frac{d^{2} u}{d x^{2}}=\frac{-2 c \sqrt{\frac{c}{a}}(+)+0}{(+)}=(-)\] [Note that $a-c x^{2}$ is zero when $x=\sqrt{\frac{c}{a}}$ ]

Therefore, $x=\sqrt{\dfrac{c}{a}}$ makes $u$ a maximum.

$x$ cannot be negative (What is the meaning of negative output of an electric generator?), but even if $x<0$ were acceptable, when $x=-\sqrt{\frac{c}{a}}$, we have

\[\frac{d^{2} u}{d x^{2}}=\frac{-2 c\left(-\sqrt{\frac{c}{a}}\right)(+)+0}{(+)}=(+),\] Therefore $u$ is a minimum when $x=-\sqrt{\dfrac{c}{a}}$.

Exercise 12.10. Suppose it to be known that consumption of coal by a certain steamer may be represented by the formula $y = 0.3 + 0.001v^3$; where $y$ is the number of tons of coal burned per hour and $v$ is the speed expressed in nautical miles per hour. The cost of wages, interest on capital, and depreciation of that ship are together equal, per hour, to the cost of $1$ ton of coal. What speed will make the total cost of a voyage of $1000$ nautical miles a minimum? And, if coal costs $\$132$ per ton, what will that minimum cost of the voyage amount to?

Answer

Speed $\sqrt[3]{650}\approx 8.66$ nautical miles per hour. Time taken $115.44$ hours.
Minimum cost $\$29714.7$.

Solution

\[\frac{\text{no. of tons}}{\text{hr}}=y=0.3+0.001 v^3\] Since the cost of other expenses is equivalent to $1\ \frac{\text{ton of coal}}{\text{hr}}$, if the voyage takes $t$ hours, then the total cost of the voyage is \[\text{cost }= a\left(y\cdot t+t\right)=a(y+1)t\] where $a$ is the cost of coal per ton.

If the speed of the steamer is $v$, since the voyage is $1000$ nautical miles, the time of the voyage is \[t=\frac{1000}{v}\] Therefore, we can write that the cost of the voyage is \[\begin{align} \text{cost } &=a\left(1.3+0.001v^3\right)\frac{1000}{v}\\ &=a\left(\frac{1300}{v}+v^2\right). \end{align}\]

To minimize the cost, we differentiate cost with respect to velocity and set the result equal to zero:

\[\frac{d \text{ cost}}{dv}=a\left(-\frac{1300}{v^2}+2v\right)=0\]

\[\Rightarrow 2v=\frac{1300}{v^2}\] \[\Rightarrow v=\sqrt[3]{650}\approx 8.662\]

8.662 nautical miles per hour is the speed that will make the total cost a minimum.

If the steamer moves by the speed $\sqrt[3]{650}$, the voyage takes $\dfrac{1000}{\sqrt[3]{650}}\approx 115.442$ hours.

To find the minimum cost, we have to calculate $\text{cost} = a\left(\dfrac{1300}{v}+v^2\right)$ for $a=132$ and $v=\sqrt[3]{650}\approx 8.66$: \[\text{cost} \approx 132\left(\frac{1300}{8.662}+8.662^2 \right)\approx 29714.7.\]

Exercise 12.11. Find the maxima and minima of\[y = \pm\frac{x}{6}\sqrt{x(10-x)}.\]

Answer

Max. and min. for $x = 7.5$, $y \approx \pm 5.413$.

Solution

First consider \[y=\frac{x}{6} \sqrt{x(10-x)}\] Then \[\frac{d y}{d x}=\frac{1}{6} \sqrt{x(10-x)}+\frac{x}{6} \frac{d\left\{ \sqrt{x(10-x)} \right\}}{d x}\]

To differentiate $\sqrt{x(10-x)}$, let $u=x(10-x)$ and \[\begin{align} \frac{d \left(\sqrt{u}\right)}{d x} & =\frac{d \left(\sqrt{u}\right)}{d u} \cdot \frac{d u}{d x} \\ & =\frac{1}{2 \sqrt{u}} \cdot(10-2 x) \\ & =\frac{1}{\sqrt{x(10-x)}} \cdot(5-x) \end{align}\] Therefore,

\[\begin{align} \frac{d y}{d x}&=\frac{1}{6} \sqrt{x(10-x)}+\frac{x}{6} \frac{5-x}{\sqrt{x(10-x)}} \\ &=\frac{x(10-x)+x(5-x)}{6 \sqrt{x(10-x)}} \\ &=\frac{15 x-2 x^{2}}{6 \sqrt{x(10-x)}} \end{align}\] \[\begin{align} \frac{d y}{d x}=0 \quad &\Leftrightarrow\quad x(15-2 x)=0 \\ &\Leftrightarrow \quad x=0 \text { or } x=7.5 \end{align}\]

To distinguish between a maximum or a minimum, we need the sign of the second derivative for $x=0$ and for $x=7.5$.

\[\frac{d^{2} y}{d x^{2}}=\frac{1}{6} \frac{(15-4 x) \sqrt{x(10-x)}-\frac{d(\sqrt{x(10-x)})}{d x} \cdot\left(15 x-2 x^{2}\right)}{(\sqrt{x(10-x)})^{2}}\]

Note that, to find the sign of $\frac{d^{2} y}{d x^{2}}$, we do not need to write the expression for $\frac{d(\sqrt{x(10-x)})}{d x}$ because the result will be multiplied by $\left(15 x-2 x^{2}\right)$ which is zero for both $x=0$ and $x=7.5$.

When $x=0$ \[\frac{d^{2} y}{d x^{2}}=\frac{1}{6} \frac{15 \cdot \sqrt{10}-0}{10}>0 .\] Therefore, $x=0$ corresponds to a minimum $y=0$.

When $x=7.5$ \[\frac{d^{2} y}{d x^{2}}=\frac{1}{6} \frac{(15-4 \times 7.5) \sqrt{7.5 \times 2.5}}{7.5 \times 2.5}<0\] Hence $x=7.5$ corresponds to a maximum \[y=\frac{7.5}{6} \sqrt{7.5 \times(10-7.5)} \approx 5.413\]

If we are careless, we might say that $x=0$ correspom to a minimum $y=0$, but we notice that $y=\dfrac{x}{6} \sqrt{x(10-x)}$ is $>0$ if $x>0$ (it is not defined for $x<0$ ). However, this curve has another branch $y=-\dfrac{x}{6} \sqrt{x(10-x)}$ which is $<0$ when $x>0$. Therefore, the curve has neither a minimum nor a maximum if $x=0$.

If we consider $y=-\dfrac{x}{6} \sqrt{x(10-x)}$, its derivative is also zero when $x=0$ or $x=7.5$. The second derivative has the opposite sign of the second derivative of the other branch $y=\dfrac{x}{6} \sqrt{x(10-x)}$. Therefore, $\dfrac{d^{2} y}{d x^{2}}>0$ when $x=7.5$. Hence $y=-\dfrac{x}{6} \sqrt{x(10-x)}$ has a minimum value of $y \approx-5.413$ when $x=7.5$.

The curve $y= \pm \dfrac{x}{6} \sqrt{x(10-x)}$ is shown below.

Exercise 12.12. Find the maxima and minima of \[y= 4x^3 - x^2 - 2x + 1.\]

Answer

Min.: $x = \frac{1}{2}$, $y= 0.25$; max.: $x = - \frac{1}{3}$, $y\approx 1.407$.

Solution

\[\begin{align} y & =4 x^{3}-x^{2}-2 x+1 \\ \frac{d y}{d x} & =12 x^{2}-2 x-2=2\left(6 x^{2}-x-1\right) \\ \frac{d y}{d x}=0 &\quad \Leftrightarrow \quad x=\frac{1 \pm \sqrt{1+24}}{12}=\frac{1 \pm 5}{12} \\ \frac{d y}{d x} =0 & \quad \Leftrightarrow \quad x=\frac{1}{2} \quad \text { or } \quad x=-\frac{1}{3} \\ \frac{d^{2} y}{d x^{2}} & =2(12 x-1) \end{align}\]

When $x=\dfrac{1}{2}$, $\dfrac{d^{2} y}{d x^{2}}=10>0$, the curve is concave upward and $y$ has a minimum value of $4 \times\left(\dfrac{1}{2}\right)^{3}-\left(\dfrac{1}{2}\right)^{2}-2\left(\dfrac{1}{2}\right)+1=\dfrac{1}{4}$ when $x=\dfrac{1}{2}$.

When $x=-\dfrac{1}{3}$, $\dfrac{d^{2} y}{d x^{2}}=-10<0$, the curve is concave downward and $y$ has a maximum value of $\dfrac{38}{27} \approx 1.407$ when $x=-\dfrac{1}{3}$.

The graph of $y= 4x^3 - x^2 - 2x + 1$ is shown below.

Exercises

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