Maxima and Minima

Excerpts

• Maxima and Minima

  So, writing $\dfrac{dy}{dx} = 0$ does *not* mean that it always is $=0$; but you write it down *as a condition* in order to see how much $x$ will come out if $\dfrac{dy}{dx}$ is to be zero.

• Maxima and Minima

  It does not of itself discriminate; it finds for you the right value of $x$ but leaves you to find out for yourselves whether the corresponding $y$ is a maximum or a minimum.

• Maxima and Minima

  It is necessary therefore always to check by taking one value on either side.

• Maxima and Minima

  Let the number to be cut into two parts be called $n$. Then if $x$ is one part, the other will be $n-x$, and the product will be $x(n-x)$ or $nx-x^2$. So we write $y=nx-x^2$.

• Maxima and Minima

  This is a very useful rule, and applies to any number of factors, so that if $m+n+p=$ a constant number, $m×n×p$ is a maximum when $m=n=p$.

• Maxima and Minima

  One of the principal uses of the process of differentiating is to find out under what conditions the value of the thing differentiated becomes a maximum, or a minimum. This is often exceedingly important in engineering questions, where it is most desirable to know what conditions will make the cost of working a minimum, or will make the efficiency a maximum.

• Maxima and Minima

  Now it may sound like juggling to be assured that there is a way by which one can arrive straight at a maximum (or minimum) value without making a lot of preliminary trials or guesses.

• Maxima and Minima

  When there is put before you an equation, and you want to find that value of $x$ that will make its $y$ a minimum (or a maximum), *first differentiate it*, and having done so, write its $\dfrac{dy}{dx}$ as *equal to zero*, and then solve for $x$. Put this particular value of $x$ into the original equation, and you will then get the required value of $y$. This process is commonly called “equating to zero.”

• Maxima and Minima

  Ordinarily you are dealing with equations that are true in themselves, but, on occasions, of which the present are examples, you have to write down equations that are not necessarily true, but are only true if certain conditions are to be fulfilled; and you write them down in order, by solving them, to find the conditions which make them true.

• Maxima and Minima

  Quite so. It does not of itself discriminate; it finds for you the right value of $x$ but leaves you to find out for yourselves whether the corresponding $y$ is a maximum or a minimum. Of course, if you have plotted the curve, you know already which it will be.

• Maxima and Minima

  So now we *know* that whatever number $n$ may be, we must divide it into two equal parts if the product of the parts is to be a maximum; and the value of that maximum product will always be $ = \tfrac{1}{4} n^2$.

• Maxima and Minima

  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.

Equations

• Maxima and Minima

  y = x^2 - 4x + 7.

  The example function whose curve has a minimum, used to introduce the method.

• Maxima and Minima

  y = 3x - x^2

  The second example function, whose curve has a maximum between x = 1 and x = 2.

• Maxima and Minima

  \dfrac{dy}{dx} = 2x - 4

  The derivative of y = x^2 - 4x + 7 with respect to x.

• Maxima and Minima

  2x - 4 = 0

  The condition obtained by equating the derivative to zero, whose solution gives the stationary value of x.

• Maxima and Minima

  \frac{dy}{dx} = 0

  The condition that the curve is neither rising nor falling, written as an equation of condition to be solved for x.

• Maxima and Minima

  y = 4x + \frac{1}{x}.

  An example function used to show that equating the derivative to zero does not say whether the stationary value is a maximum or a minimum.

• Maxima and Minima

  y = nx - x^2

  The product y of the two parts when a number n is cut into parts x and n - x.

• Maxima and Minima

  \dfrac{n}{2} = x

  The value of one part for which the product of the two parts is a maximum, namely half the number.

• Maxima and Minima

  x^2 - 4x +3 = 0

  The quadratic obtained by equating dy/dx to zero for y = (1/3)x^3 - 2x^2 + 3x + 1, whose two roots give the maximum and minimum.

• Maxima and Minima

  y =\tfrac{1}{3} x^3 - 2x^2 + 3x + 1.

  The cubic function in the further examples, which has both a maximum and a minimum.

• Maxima and Minima

  (y-b)^2 + (x-a)^2 = r^2.

  The equation of a circle of radius r with centre at (a, b).

• Maxima and Minima

  y = \sqrt{r^2-(x-a)^2} + b.

  The upper-branch form of the circle's equation, solved for y.

• Maxima and Minima

  \frac{a-x}{\sqrt{r^2-(x-a)^2}} = 0.

  The condition for the circle's height y to be a maximum or minimum, which holds only at x = a.

• Maxima and Minima

  3ax^2 + b = 0

  The condition from equating dy/dx to zero for y = ax^3 + bx + c; it gives no real x when a and b have the same sign, so y has no maximum or minimum.

• Maxima and Minima

  \text{the other side} = \sqrt{(\text{diagonal})^2 - x^2}

  The side of a rectangle inscribed in a circle, found from its diagonal, which is a diameter of the circle.

• Maxima and Minima

  S = x\sqrt{4R^2 - x^2}

  The area of a rectangle inscribed in a circle of radius R, with one side x.

• Maxima and Minima

  4R^2 - 2x^2 = 0

  The condition from equating dS/dx to zero for the inscribed rectangle.

• Maxima and Minima

  x = R\sqrt{2}

  The side of the maximum-area inscribed rectangle, which makes it a square.

• Maxima and Minima

  H = \sqrt{l^2 - R^2}

  The height of a cone in terms of its slant length l and base radius R.

• Maxima and Minima

  V = \pi R^2 × \dfrac{H}{3} = \pi R^2 × \dfrac{\sqrt{l^2 - R^2}}{3}

  The volume of a cone of base radius R and height H, with H expressed through l and R.

• Maxima and Minima

  2\pi R(l^2 - R^2) - \pi R^2 = 0

  The condition printed in the cone problem after dV/dR is set to zero. FLAG: this does not follow from the line before it, which has numerator 2\pi R(l^2 - R^2) - \pi R^3; the book's printed R^2 appears to be a typo for R^3, and the stated result R = l\sqrt{2/3} follows only from the R^3 version.

• Maxima and Minima

  R = l\sqrt{\tfrac{2}{3}}

  The base radius of the cone of maximum volume for a given slant length l.

• Maxima and Minima

  y = \dfrac{x}{4-x} + \dfrac{4-x}{x}

  The function of the third worked example, which has a single minimum at x = 2.

• Maxima and Minima

  \dfrac{4}{(4-x)^2} - \dfrac{4}{x^2} = 0

  The condition from equating dy/dx to zero for the third example, which gives x = 2.

• Maxima and Minima

  y = \sqrt{1+x} + \sqrt{1-x}

  The function of the fourth worked example, which has a maximum at x = 0.

• Maxima and Minima

  \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{1+x}} - \dfrac{1}{2\sqrt{1-x}} = 0

  The derivative of the fourth example set equal to zero for maximum or minimum.

• Maxima and Minima

  \sqrt{1+x} = \sqrt{1-x}

  The condition whose only solution is x = 0 in the fourth example.

• Maxima and Minima

  y = \dfrac{x^2-5}{2x-4}

  The function of the fifth worked example, which has neither a maximum nor a minimum.

• Maxima and Minima

  x^2 - 4x + 5 = 0

  The quadratic obtained from dy/dx = 0 in the fifth example; its roots are 2 ± i, not the 5/2 ± i printed in the next line.

• Maxima and Minima

  x = \tfrac{5}{2} ± \sqrt{-1}

  FLAG: the book's printed roots of x^2 - 4x + 5 = 0 are wrong; completing the square gives x = 2 ± i. The conclusion that there are no real roots, hence no maximum or minimum, is unaffected.

• Maxima and Minima

  (y-x^2)^2 = x^5

  The implicit curve of the sixth example, which splits into two branches y = x^2 ± x^(5/2).

• Maxima and Minima

  y = x^2 ± x^{\efrac{5}{2}}

  The sixth example's curve written as two explicit functions of x.

• Maxima and Minima

  \dfrac{dy}{dx} = 2x ± \tfrac{5}{2} x^{\efrac{3}{2}} = 0

  The derivative of the two branches set equal to zero for a maximum or minimum.

• Maxima and Minima

  2 ± \tfrac{5}{2} x^{\efrac{1}{2}} = 0

  The condition on the non-zero stationary point of the sixth example.

• Maxima and Minima

  x = \tfrac{16}{25}

  The second stationary value of x for the sixth example, obtained from the non-zero condition.

• Maxima and Minima

  S = 2(\pi r^2)+ 2 \pi r × 2r = 6 \pi r^2

  The total surface area of a cylinder whose height is twice the radius r of its base.

• Maxima and Minima

  V = \pi r^2 × 2r=2 \pi r^3

  The volume of a cylinder whose height is twice the radius r of its base.

• Maxima and Minima

  \frac{dS}{dr} = 12\pi r

  The rate of change of the surface area with the base radius r.

• Maxima and Minima

  \frac{dV}{dr}=6 \pi r^2

  The rate of change of the volume with the base radius r.

Problems