Maxima and Minima

Example 11.1. Find the minimum value of \(y\) if \[y = x^2 - 4x + 7.\]

Solution. Differentiating, we get: \[\dfrac{dy}{dx} = 2x - 4.\] Now equate this to zero, thus: \[2x - 4 = 0.\] Solving this equation for \(x\), we get: \[\begin{align} 2x &= 4, \\ x &= 2. \end{align}\]

Now, we know that the maximum (or minimum) will occur exactly when \(x=2\).

Putting the value \(x=2\) into the original equation, we get \[\begin{align} y &= 2^2 - (4\times2) + 7 \\ &= 4 - 8 + 7 \\ &= 3. \end{align}\]

Now look back at Fig. 11.1, and you will see that the minimum occurs when \(x = 2\), and that this minimum of \(y = 3\).

Example 11.2. Find the maximum value of \(y\) if \[y = 3x - x^2.\]

Solution. Differentiating, \[\frac{dy}{dx} = 3 - 2x.\] Equating to zero, \[3 - 2x = 0,\] whence \[x = 1 \frac{1}{2};\] and putting this value of \(x\) into the original equation, we find: \[\begin{align} y &= 4 \frac{1}{2} - \left(1 \frac{1}{2} \times 1 \frac{1}{2}\right), \\ y &= 2 \frac{1}{4}. \end{align}\] This gives us exactly the information as to which the method of trying a lot of values left us uncertain.

Example 11.3. Let \[\begin{align} y &= x^2 - x; \end{align}\] and let us find whether this function has a maximum or minimum; and if so, test whether it is a maximum or a minimum.

Solution. Differentiating, we get \[\frac{dy}{dx} = 2x - 1.\] Equating to zero, we get \[2x - 1 = 0,\] whence \[2x = 1,\] or \[x = \frac{1}{2}.\]

That is to say, when \(x\) is made \(=\frac{1}{2}\), the corresponding value of \(y\) will be either a maximum or a minimum. Accordingly, putting \(x=\frac{1}{2}\) in the original equation, we get \[\begin{align} y &= \left(\frac{1}{2}\right)^2 - \frac{1}{2}, \\ \end{align}\] or \[\begin{align} y &= -\frac{1}{4}. \end{align}\]

Is this a maximum or a minimum? To test it, try putting \(x\) a little bigger than \(\frac{1}{2}\),—say make \(x=0.6\). Then \[y = (0.6)^2 - 0.6 = 0.36 - 0.6 = -0.24,\] which is higher up than \(-0.25\); showing that \(y = -0.25\) is a minimum.

The curve is plotted below. As we can see, the minimum of \(y\) is \(-\frac{1}{4}\), which occurs when \(x=\frac{1}{2}\).

Example 11.4. Consider a curve whose equation is: \[y =\frac{1}{3} x^3 - 2x^2 + 3x + 1.\] Now \[\dfrac{dy}{dx} = x^2 - 4x +3.\]

Equating to zero, we get the quadratic, \[x^2 - 4x +3 = 0;\] and solving the quadratic gives us two roots, viz. \[\left\{\begin{align} x &= 3 \\ x &= 1. \end{align}\right.\]

Now, when \(x=3\), \(y=1\); and when \(x=1\), \(y=2\frac{1}{3}\). The first of these is a minimum, the second a maximum.

The curve itself may be plotted (as in the following figure) from the values calculated, as below, from the original equation.

Example 11.5. The equation to a circle of radius \(r\) (\(r\geq 0\)), having its centre \(C\) at the point whose coordinates are \(x=a\), \(y=b\), as depicted below, is: \[(y-b)^2 + (x-a)^2 = r^2.\]

This may be transformed into \[y = \pm\sqrt{r^2-(x-a)^2} + b.\]

Now we know beforehand, by mere inspection of the figure, that when \(x=a\), \(y\) will be either at its maximum value, \(b+r\), or else at its minimum value, \(b-r\). But let us not take advantage of this knowledge; let us set about finding what value of \(x\) will make \(y\) a maximum or a minimum, by the process of differentiating and equating to zero. \[\begin{align} \frac{dy}{dx} &=\pm \frac{1}{2} \frac{1}{\sqrt{r^2-(x-a)^2}} \times (2a-2x), \end{align}\] which reduces to \[\begin{align} \frac{dy}{dx} &=\pm \frac{a-x}{\sqrt{r^2-(x-a)^2}}. \end{align}\]

Then the condition for \(y\) being maximum or minimum is: \[\frac{a-x}{\sqrt{r^2-(x-a)^2}} = 0.\]

Since no value whatever of \(x\) will make the denominator infinite, the only condition to give zero is \[x = a.\] Inserting this value in the original equation for the circle, we find \[y = \pm \sqrt{r^2}+b;\] and as \(\sqrt{r^2}=|r|=r\), we have two resulting values of \(y\), \[\begin{cases} y=b+r \\ y=b-r. \end{cases}\]

The first of these is the maximum, at the top; the second the minimum, at the bottom.

Example 11.6. Let \[y = ax^3 + bx + c.\qquad(a,b>0)\]

Then \[\frac{dy}{dx} = 3ax^2 + b.\]

Equating this to zero, we get \(3ax^2 + b = 0\), \[x^2 = \frac{-b}{3a},\] and \[x = \sqrt{\frac{-b}{3a}}\quad \text{ or }\quad x=-\sqrt{\frac{-b}{3a}}\]

which is impossible.¹ Therefore \(y\) has no maximum nor minimum.

The following figure depicts the graph of \(y=ax^3+bx+c\) with specific values for \(a=\frac{1}{5}\), \(b=\frac{1}{7}\), and \(c=-1\). It is evident from the figure that the function \(y\) does not possess any maximum or minimum points.

Example 11.7. What are the sides of the rectangle of maximum area inscribed in a circle of radius \(R\)?

Solution. If one side is called \(x\), \[\text{the other side} = \sqrt{(\text{diagonal})^2 - x^2};\] and as the diagonal of the rectangle is necessarily a diameter, the other side \(= \sqrt{4R^2 - x^2}\) (see the following figure).

Then, area of rectangle \(S = x\sqrt{4R^2 - x^2}\), \[\frac{dS}{dx} = x \times \dfrac{d\left(\sqrt{4R^2 - x^2}\,\right)}{dx} + \sqrt{4R^2 - x^2} \times \dfrac{d(x)}{dx}.\tag{Use Product Rule}\]

If you have forgotten how to differentiate \(\sqrt{4R^2-x^2}\), here is a hint: write \(4R^2-x^2=w\) and \(y=\sqrt{w}\), and seek \(\dfrac{dy}{dw}\) and \(\dfrac{dw}{dx}\); fight it out, and only if you can’t get on refer to the Chain Rule on page .

You will get \[\begin{align} \dfrac{dS}{dx} &= x \times \left( -\dfrac{x}{\sqrt{4R^2 - x^2}} \right)+ \sqrt{4R^2 - x^2}\\ &= \dfrac{4R^2 - 2x^2}{\sqrt{4R^2 - x^2}}. \end{align}\]

For maximum or minimum we must have \[\dfrac{4R^2 - 2x^2}{\sqrt{4R^2 - x^2}} = 0;\] that is, \(4R^2 - 2x^2 = 0\) and \(x = R\sqrt{2}\).

The other side \({} = \sqrt{4R^2 - 2R^2} = R\sqrt{2}\); the two sides are equal; the figure is a square the side of which is equal to the diagonal of the square constructed on the radius. In this case it is, of course, a maximum with which we are dealing.

Example 11.8. What is the radius of the opening of a conical vessel the sloping side of which has a length \(l\) when the capacity of the vessel is greatest?

.

Solution. If \(R\) be the radius and \(H\) the corresponding height, \(H = \sqrt{l^2 - R^2}\) (see the following figure). \[\text{Volume } V = \pi R^2 \times \dfrac{H}{3} = \pi R^2 \times \dfrac{\sqrt{l^2 - R^2}}{3}.\]

Proceeding as in the previous problem, we get \[\begin{align} \dfrac{dV}{dR} &= \pi R^2 \times \left(-\dfrac{R}{3\sqrt{l^2 - R^2}}\right) + \dfrac{2\pi R}{3} \sqrt{l^2 - R^2} \\ &= \dfrac{2\pi R(l^2 - R^2) - \pi R^3}{3\sqrt{l^2 - R^2}} = 0 \end{align}\] for maximum or minimum.

Or, \(2\pi R(l^2 - R^2) - \pi R^2 = 0\), and \(R = l\sqrt{\dfrac{2}{3}}\), for a maximum, obviously.

Example 11.9. Find the maxima and minima of the function \[y = \dfrac{x}{4-x} + \dfrac{4-x}{x}.\] Solution. We get \[\dfrac{dy}{dx} = \dfrac{(4-x)-(-x)}{(4-x)^2} + \dfrac{-x - (4-x)}{x^2} = 0\] for maximum or minimum; or \[\dfrac{4}{(4-x)^2} - \dfrac{4}{x^2} = 0 \quad\text{and}\quad x = 2.\]

There is only one value, hence only one maximum or minimum. \[\begin{align} \text{For}\quad x &= 2,\phantom{.5}\quad y = 2, \\ \text{for}\quad x &= 1.5,\quad y = 2.27, \\ \text{for}\quad x &= 2.5,\quad y = 2.27; \end{align}\] it is therefore a minimum.

The graph of \(y=\dfrac{x}{4-x} + \dfrac{4-x}{x}\) is shown below.

Example 11.10. Find the maxima and minima of the function \(y = \sqrt{1+x} + \sqrt{1-x}\).

Solution. Differentiating gives at once (see example 9.1) \[\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{1+x}} - \dfrac{1}{2\sqrt{1-x}} = 0\] for maximum or minimum.

Hence \(\sqrt{1+x} = \sqrt{1-x}\) and \(x = 0\), the only solution

For \(x=0\),\(y=2\).

For \(x=\pm0.5\), \(y= \sqrt{1.5}+\sqrt{0.5}\approx1.932\), so this is a maximum.

The graph of \(y=\sqrt{1+x} + \sqrt{1-x}\) is shown below.

Example 11.11. Find the maxima and minima of the function \[y = \dfrac{x^2-5}{2x-4}.\]

Solution. We have \[\dfrac{dy}{dx} = \dfrac{(2x-4) \times 2x - (x^2-5)2}{(2x-4)^2} = 0\] for maximum or minimum; or \[\dfrac{2x^2 - 8x + 10}{{(2x - 4)^2}} = 0;\] or \(x^2 - 4x + 5 = 0\); which has for solutions \[x = \frac{5}{2} \pm \sqrt{-1}.\]

These being imaginary, there is no real value of \(x\) for which \(\dfrac{dy}{dx} = 0\); hence there is neither maximum nor minimum.

The following figure shows the graph of \(y=\dfrac{x^2-5}{2x-4}\).

Example 11.12. Find the maxima and minima of the function \[(y-x^2)^2 = x^5.\]

Solution. This may be written \(y = x^2 \pm x^{\frac{5}{2}}\). \[\dfrac{dy}{dx} = 2x \pm \frac{5}{2} x^{\frac{3}{2}} = 0 \quad\text{for maximum or minimum};\] that is, \(x(2 \pm \frac{5}{2} x^{\frac{1}{2}}) = 0\), which is satisfied for \(x = 0\), and for \(2 \pm \frac{5}{2} x^{\frac{1}{2}} = 0\), that is for \(x=\frac{16}{25}\). So there are two solutions.

Taking first \(x = 0\). If \(x = -0.5\), \(y = 0.25 \pm \sqrt{-(0.5)^5}\), and if \(x = +0.5\), \(y = 0.25 \pm \sqrt{{(0.5)^5}}\). On one side \(y\) is imaginary; that is, there is no value of \(y\) that can be represented by a graph; the latter is therefore entirely on the right side of the axis of \(y\) (see the following figure).

On plotting the graph it will be found that the curve goes to the origin, as if there were a minimum there; but instead of continuing beyond, as it should do for a minimum, it retraces its steps (forming what is called a “cusp”). There is no minimum, therefore, although the condition for a minimum is satisfied, namely \(\dfrac{dy}{dx} = 0\). It is necessary therefore always to check by taking one value on either side.

Now, if we take \(x = \frac{16}{25} = 0.64\). If \(x = 0.64\), \(y = 0.7373\) and \(y = 0.0819\); if \(x = 0.6\), \(y\) becomes \(0.6389\) and \(0.0811\); and if \(x = 0.7\), \(y\) becomes \(0.8996\) and \(0.0804\).

This shows that there are two branches of the curve; the upper one does not pass through a maximum, but the lower one does.

Example 11.13. Find the maxima and minima of the function \[y=\dfrac{(x-1)^2}{(x+1)^3}.\]

Solution. First, we find the derivative of \(y\) \[\frac{dy}{dx}=\frac{2(x-1)(x+1)^3-3(x+1)^2(x-1)^2}{(x+1)^6}\tag{Quotient Rule}\] or \[\begin{align} \frac{dy}{dx}&=\frac{2(x-1)(x+1)-3(x-1)^2}{(x+1)^4}\\ &=\frac{(x-1)[2(x+1)-3(x-1)]}{(x+1)^4}\\ &=\frac{(x-1)(5-x)}{(x+1)^4} \end{align}\] Therefore, for \(x=1\) and \(x=5\), we have \(\dfrac{dy}{dx}=0\).

To determine whether each of these values of \(x\) gives a maximum or minimum, we can apply the First Derivative Test. Considering the sign of the derivative for values of \(x\) that are slightly less than \(1\) and slightly larger than \(1\), we have:

\[\left.\begin{align} \text{When }x<1, \ \dfrac{dy}{dx}=\frac{(-)(+)}{+}=-\\ \text{When }x>1, \ \dfrac{dy}{dx}=\frac{(+)(+)}{+}=+ \end{align}\right\}\] Hence \(x=1\) corresponds to a minimum \(y=0\).

Similarly, considering the sign of the derivative for the values of \(x\) that are slightly less than \(5\) and slightly greater than \(5\), we get: \[\left.\begin{align} \text{When }x<5, \ \dfrac{dy}{dx}=\frac{(+)(+)}{+}=+\\ \text{When }x>5, \ \dfrac{dy}{dx}=\frac{(+)(-)}{+}=- \end{align}\right\}\] Hence \(x=5\) corresponds to a maximum \(y=\frac{2}{27}\).

This curve is plotted below.