Distinguishing between local maxima and minima

2 Distinguishing between local maxima and minima

We might ask if it is possible to predict when a stationary point is a local maximum, a local minimum or a point of inflection without the necessity of drawing the curve. To do this we highlight the general characteristics of curves in the neighbourhood of local maxima and minima.

For example: at a local maximum (located at $x_{0}$ say) Figure 11 describes the situation:

Figure 11

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If we draw a graph of the derivative $\frac{d f}{d x}$ against $x$ then, near a local maximum, it must take one of two basic shapes described in Figure 12:

Figure 12

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$In case (a) \frac{d}{d x} (\frac{d f}{d x}) |_{x_{0}} \equiv tan α < 0 whilst in case (b) \frac{d}{d x} (\frac{d f}{d x}) |_{x_{0}} = 0$

We reach the conclusion that at a stationary point which is a maximum the value of the second derivative $\frac{d^{2} f}{d x^{2}}$ is either negative or zero.

Near a local minimum the above graphs are inverted. Figure 13 shows a local minimum.

Figure 13

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Figure 134 shows the two possible graphs of the derivative:

Figure 14

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Here, $for case (a) \frac{d}{d x} (\frac{d f}{d x}) |_{x_{0}} = tan β > 0 whilst in (b) \frac{d}{d x} (\frac{d f}{d x}) |_{x_{0}} = 0.$

In this case we conclude that at a stationary point which is a minimum the value of the second derivative $\frac{d^{2} f}{d x^{2}}$ is either positive or zero. For the third possibility for a stationary point - a point of inflection - the graph of $f (x)$ against $x$ and of $\frac{d f}{d x}$ against $x$ take one of two forms as shown in Figure 15.

Figure 15

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$For either of these cases \frac{d}{d x} (\frac{d f}{d x}) |_{x_{0}} = 0$

The sketches and analysis of the shape of a curve $y = f (x)$ in the near neighbourhood of stationary points allow us to make the following important deduction:

Key Point 4

If $x_{0}$ locates a stationary point of $f (x)$ , so that $\frac{d f}{d x} |_{x_{0}} = 0$ , then the stationary point

is a local minimum if $\frac{d^{2} f}{d x^{2}} |_{x_{0}} > 0$

is a local maximum if $\frac{d^{2} f}{d x^{2}} |_{x_{0}} < 0$

is inconclusive if $\frac{d^{2} f}{d x^{2}} |_{x_{0}} = 0$

Example 3

Find the stationary points of the function $f (x) = x^{3} - 6 x$ .

Are these stationary points local maxima or local minima?

Solution

$\frac{d f}{d x} = 3 x^{2} - 6$ . At a stationary point $\frac{d f}{d x} = 0$ so $3 x^{2} - 6 = 0$ , implying $x = \pm \sqrt{2}$ .

Thus $f (x)$ has stationary points at $x = \sqrt{2}$ and $x = - \sqrt{2}$ . To decide if these are maxima or minima we examine the value of the second derivative of $f (x)$ at the stationary points.

$\frac{d^{2} f}{d x^{2}} = 6 x$ so $\frac{d^{2} f}{d x^{2}} |_{x = \sqrt{2}} = 6 \sqrt{2} > 0$ . Hence $x = \sqrt{2}$ locates a local minimum.

Similarly $\frac{d^{2} f}{d x^{2}} |_{x = - \sqrt{2}} = - 6 \sqrt{2} < 0$ . Hence $x = - \sqrt{2}$ locates a local maximum.

A sketch of the curve confirms this analysis:

Figure 16

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Task!

For the function $f (x) = cos 2 x, 0.1 \leq x \leq 6$ , find the positions of any local minima or maxima and distinguish between them.

Calculate the first derivative and locate stationary points:

$\frac{d f}{d x} = - 2 sin 2 x$ .

Hence stationary points are at values of $x$ in the range specified for which $sin 2 x = 0$ i.e. at $2 x = π$ or $2 x = 2 π$ or $2 x = 3 π$ (making sure $x$ is within the range $0.1 \leq x \leq 6$ )

$∴$ Stationary points at $x = \frac{π}{2}, x = π, x = \frac{3 π}{2}$

Now calculate the second derivative:

$\frac{d^{2} f}{d x^{2}} = - 4 cos 2 x$

Finally: evaluate the second derivative at each stationary points and draw appropriate conclusions:

$\frac{d^{2} f}{d x^{2}} |_{x = \frac{π}{2}} = - 4 cos π = 4 > 0$ $∴ x = \frac{π}{2}$ locates a local minimum.

$\frac{d^{2} f}{d x^{2}} |_{x = π} = - 4 cos 2 π = - 4 < 0 ∴ x = π$ locates a local maximum.

$\frac{d^{2} f}{d x^{2}} |_{x = \frac{3 π}{2}} = - 4 cos 3 π = 4 > 0 ∴ x = \frac{3 π}{2}$ locates a local minimum.

Task!

Determine the local maxima and/or minima of the function $y = x^{4} - \frac{1}{3} x^{3}$

First obtain the positions of the stationary points:

$\frac{d f}{d x} = 4 x^{3} - x^{2} = x^{2} (4 x - 1)$ $\frac{d f}{d x} = 0$ when $x = 0$ or when $x = 1 ∕ 4$

Now obtain the value of the second derivatives at the stationary points:

$\frac{d^{2} f}{d x^{2}} = 12 x^{2} - 2 x$ $\frac{d^{2} f}{d x^{2}} |_{x = 0} = 0, which is inconclusive.$

$\frac{d^{2} f}{d x^{2}} |_{x = 1 ∕ 4} = \frac{12}{16} - \frac{1}{2} = \frac{1}{4} > 0$ Hence $x = \frac{1}{4}$ locates a local minimum.

Using this analysis we cannot decide whether the stationary point at $x = 0$ is a local maximum, minimum or a point of inflection. However, just to the left of $x = 0$ the value of $\frac{d f}{d x}$ (which equals $x^{2} (4 x - 1)$ ) is negative whilst just to the right of $x = 0$ the value of $\frac{d f}{d x}$ is negative again. Hence the stationary point at $x = 0$ is a point of inflection . This is confirmed by sketching the curve as in Figure 17.

Figure 17

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Task!

A materials store is to be constructed next to a 3 metre high stone wall (shown as $O A$ in the cross section in the diagram). The roof $(A B)$ and front $(B C)$ are to be constructed from corrugated metal sheeting. Only 6 metre length sheets are available. Each of them is to be cut into two parts such that one part is used for the roof and the other is used for the front. Find the dimensions $x, y$ of the store that result in the maximum cross-sectional area. Hence determine the maximum cross-sectional area.

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Note that the store has the shape of a trapezium. So the cross-sectional area $(A)$ of the store is given by the formula: Area $=$ average length of parallel sides $\times$ distance between parallel sides:

$A = \frac{1}{2} (y + 3) x$ (1)

The lengths $x$ and $y$ are related through the fact that $A B + B C = 6,$ where $B C = y$ and $A B = \sqrt{x^{2} + {(3 - y)}^{2}}$ . Hence $\sqrt{x^{2} + {(3 - y)}^{2}} + y = 6$ . This equation can be rearranged in the following way:

$\sqrt{x^{2} + {(3 - y)}^{2}} = 6 - y \Leftrightarrow x^{2} + {(3 - y)}^{2} = {(6 - y)}^{2}$ i.e. $x^{2} + 9 - 6 y + y^{2} = 36 - 12 y + y^{2}$

which implies that $x^{2} + 6 y = 27$ (2)

It is necessary to eliminate either $x$ or $y$ from (1) and (2) to obtain an equation in a single variable. Using $y$ instead of $x$ as the variable will avoid having square roots appearing in the expression for the cross-sectional area. Hence from Equation (2)

$y = \frac{27 - x^{2}}{6}$ (3)

Substituting for $y$ from Equation (3) into Equation (1) gives

$A = \frac{1}{2} (\frac{27 - x^{2}}{6} + 3) x = \frac{1}{2} (\frac{27 - x^{2} + 18}{6}) x = \frac{1}{12} (45 x - x^{3})$ (4)

To find turning points, we evaluate $\frac{d A}{d x}$ from Equation (4) to get

$\frac{d A}{d x} = \frac{1}{12} (45 - 3 x^{2})$ (5)

Solving the equation $\frac{d A}{d x} = 0$ gives $\frac{1}{12} (45 - 3 x^{2}) = 0 \Rightarrow 45 - 3 x^{2} = 0$

Hence $x = \pm \sqrt{15} = \pm 3.8730$ . Only $x > 0$ is of interest, so

$x = \sqrt{15} = 3.8730$ 6 (6)

gives the required turning point.

Check: Differentiating Equation (5) and using the positive $x$ solution (6) gives

$\frac{d^{2} A}{d x^{2}} = - \frac{6 x}{12} = - \frac{x}{2} = - \frac{3.8730}{2} < 0$

Since the second derivative is negative then the cross-sectional area is a maximum. This is the only turning point identified for $A > 0$ and it is identified as a maximum. To find the corresponding value of $y$ , substitute $x = 3.8730$ into Equation (3) to get $y = \frac{27 - 3.873 0^{2}}{6} = 2.0000$

So the values of $x$ and $y$ that yield the maximum cross-sectional area are 3.8730 m and 2.00000 m respectively. To find the maximum cross-sectional area, substitute for $x = 3.8730$ into Equation (5) to get

$A = \frac{1}{2} (45 \times 3.8730 - 3.873 0^{3}) = 9.6825$

So the maximum cross-sectional area of the store is $9.68 m^{2}$ to 2 d.p.

Task!

Equivalent resistance in an electrical circuit

Current distributes itself in the wires of an electrical circuit so as to minimise the total power consumption i.e. the rate at which heat is produced. The power $(p)$ dissipated in an electrical circuit is given by the product of voltage $(v)$ and current $(i)$ flowing in the circuit, i.e. $p = v i .$ The voltage across a resistor is the product of current and resistance $(r) .$ This means that the power dissipated in a resistor is given by $p = i^{2} r .$

Suppose that an electrical circuit contains three resistors $r_{1}, r_{2}, r_{3}$ and $i_{1}$ represents the current flowing through both $r_{1}$ and $r_{2}$ , and that $(i - i_{1})$ represents the current flowing through $r_{3}$ (see diagram):

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Write down an expression for the power dissipated in the circuit:

$p = i_{1}^{2} r_{1} + i_{1}^{2} r_{2} + {(i - i_{1})}^{2} r_{3}$
Show that the power dissipated is a minimum when $i_{1} = \frac{r_{3}}{r_{1} + r_{2} + r_{3}} i :$

Differentiate result (1) with respect to $i_{1}$ :
$\begin{array}{rcl} \frac{d p}{d i_{1}} & = & 2 i_{1} r_{1} + 2 i_{1} r_{2} + 2 (i - i_{1}) (- 1) r_{3} \\ = & 2 i_{1} (r_{1} + r_{2} + r_{3}) - 2 i r_{3} \end{array}$
This is zero when

$i_{1} = \frac{r_{3}}{r_{1} + r_{2} + r_{3}} i .$

To check if this represents a minimum, differentiate again:

$\frac{d^{2} p}{d i_{1}^{2}} = 2 (r_{1} + r_{2} + r_{3})$

This is positive, so the previous result represents a minimum.
If $R$ is the equivalent resistance of the circuit, i.e. of $r_{1}, r_{2}$ and $r_{3}$ , for minimum power dissipation and the corresponding voltage $V$ across the circuit is given by $V = i R = i_{1} (r_{1} + r_{2}),$ show that
$R = \frac{(r_{1} + r_{2}) r_{3}}{r_{1} + r_{2} + r_{3}} .$

Substituting for $i_{1}$ in $i R = i_{1} (r_{1} + r_{2})$ gives

$i R = \frac{r_{3} (r_{1} + r_{2})}{r_{1} + r_{2} + r_{3}} i .$

So

$R = \frac{(r_{1} + r_{2}) r_{3}}{r_{1} + r_{2} + r_{3}} .$

Note In this problem $R_{1}$ and $R_{2}$ could be replaced by a single resistor. However, treating them as separate allows the possibility of considering more general situations (variable resistors or temperature dependent resistors).

2 Distinguishing between local maxima and minima

Key Point 4

Example 3

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