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90. To find whether a proposed quantity admits of a Maximum or a Minimum.

Every algebraic expression does not admit of a maximum or minimum, properly so called; for it may either increase

negative we assign to e and f, the least value of 16+4ecf would be found by taking e and f each=0; or if 4eef were always a negative quantity, whatever positive or negative values we give to e and f, then 16+4eef would be a maximum by taking e and feach0: but if by changing the signs of è and ƒ, or their ratio, the formula 4eef can be changed from positive to negative or from negative to positive, it is clear that the quantity 16+4eef cannot be either a maximum or minimum by making e and feach =0.

Now 4e2

ef=e2 (4 and 4

e can

evidently be made positive or negative at pleasure. For example, if e=1,ƒ=1, we have 4e3 —ef = 3. In this case x=2+1=3, y=16+1=17, and therefore 4x2 —xy+2y=16+3=19. Ife=1,f=5, then 4 1 Here 2+1=3, y=16+5=21, and

4x — xÿ+ 2y = 16—1—15.

Thus we see the reason why the proposed formula cannot be a maximum or
minimum when ✰
x=2, y=16.

Let any expression in x and y be proposed to be a maximum or minimum. Find by the rule in the text the values a and b, of x and y, corresponding to the maximum or minimum. Put a=a+e, y=b+f, and by substituting a +e, and bf for x and y, in the expression proposed to be made a maximum or minimum, let ce2 + Deƒ+Eƒ2

be that part of the proposed expression which involves the second powers of e and f, C, D, and E being obtained in terms of a and b. Put u and we have

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Ce2 +Deƒ+xƒ2 ——— } 4c2u2 +4cdu+D*+4ce—d
+Def+Ef 4c

Put 4cu + D


and if 4 CE- D =D be positive, put it and we have
Ce2 + Def + Eƒ =


f3 {a2 + B2 }

It is now evident that when 4cE-2 is positive, the quantity ce+Defeƒ2
cannot be made to change its sign, but must remain of the same sign with the
coefficient C, and consequently when 4CE-D2 is a positive quantity, the proposed
expression is a maximum or minimum according as c is negative or affirmative.
When 4CE—D2 is negative, put it —— ß2, and we have


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Now since a 2cu+D, in which u is arbitrary and may be taken at pleasure, it is evident that a may become of any value whatever, while ß3 remains constant,

and therefore by taking different value

of u the quantity a
of u the quantity aa -ẞ2 may be both

positive and negative, and of course ce +Def+Eƒ may change its sign by as

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signing different values to; and therefore when 4ce D2 is negative, the proposed formula can neither be a maximum nor minimum.

In the


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continually to infinity, or decrease continually to nothing; and in both these cases there is neither a proper maximum nor minimum; for the true maximum is that finite value to which an expression increases, and after which it decreases again : and the minimum is that finite value to which the expression decreases, and after that it increases again. Therefore, when the expression admits of a maximum, its fluxion is positive before the point, and negative after it; but when it admits of a minimum, its fluxion is negative before, and positive after it. Hence then, taking the fluxion of the expression a little before the fluxion is equal to nothing, and again a little after the same, if the former fluxion be positive, and the latter negative, the middle state is a maximum, but if the former fluxion be negative, and the latter positive, the middle state is mini



So, if we would find the quantity ax x2 a maximum or minimum; make its fluxion equal to nothing, that is,

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Q, or (a — 2x)x

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2xx =
2x 0, or x=1a at that state.

dividing by x, gives Now, if in the fluxion 2x) x the value of x be taken rather less that its true valae a, that fluxion will evidently be positive; but if a be taken somewhat greater than a the value of a 2x, and consequently of the fluxion is as evidently negative. Therefore, the fluxion of axx being positive before, and negative after the state when its fluxion is = 0, it follows that at this state the expression is not a minimum but a maximum. Again, taking the expression x3 ax2, its fluxion 3x2x 2axx=(3x-2a)x 0; this divided by xx gives 3x-2a=0,


In the example in the text we have ce + Deƒ + Eƒ2 — 4e2 —ef.

·1, E≈é, hence 4ɛɛ —D

Here 4, D ——
-D2-1, and therefore the formula
4x3-2xy + 2y has neither maximum nor minimum.

If in the substitution of a +e and b+ƒ for x, and y in the given expression,
the coefficients C, D, E of the second order vanish, the increment will depend on the
terms of the third order, in which case the quantity can have neither maximum nor
minimum. Should the terms of the third order also vanish, we must have recourse
to the terms of the fourth order, the sum of which as in the case of the second
powers must be incapable of changing its sign that there may be a maximum.
In general, that the proposed formula may be a maximum or minimum, the incre-
ment must have its lowest powers of e and ƒ of an even order, and the sum of
all these even terms must be incapable of changing its sign; or, which amounts
to the same thing, the sum of these terms equated to 0, must give only imaginary
value for each of the quantities e, ƒ.

The same rule is equally applicable when there are any numbers of independ ent variables, x, y, z, &c.

The coefficient c, D, E, &c. may be obtained with the greatest ease, by taking successive fluxions according to the method of partial differentials.


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and x=2a, its true value when the fluxion of 3-ax2 is equal to nothing. But now to know whether the given expression be a maximum or a minimum at that time, take x a little less than a in the value of the fluxion (3x - 20)xx, and this will evidently be negative; and again, taking x a little more than a, the value of 3x-2a or of the fluxion, is as evidently positive. Therefore the fluxion of x-ax2 being negative before that fluxion is =0, and positive after it, it follows that in this state the quantity 3-ax2 admits of a minimum, but not of a maximum.


EXAM. 1. To divide a line, or any other given quantity a, into two parts, so that their rectangle or product may be the greatest possible.

EXAM. 2. To divide the given quantity a into two parts such that the product of the m power of one, by the n power of the other, may be a maximum.

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EXAM. 3. To divide the given quantity a into three parts such that the continual product of them all may be a maxi


EXAM. 4. To divide the given quantity a into three parts such, that the continual product of the 1st, the square of the 2d, and the cube of the 3d, may be a maximum.

EXAM. 5. To determine a fraction such that the difference between its m power and n power shall be the greatest possible.

EXAM. 6. To divide the number 80 into two such parts x and y, that 2x2+xy+3y2 may be a minimum.

EXAM. 7. To find the greatest rectangle that can be inscribed in a given right-angled triangle.

EXAM. 8. To find the greatest rectangle that can be inscribed in the quadrant of a given circle.

EXAM. 9. To find the least right-angled triangle that can circumscribe the quadrant of a given circle.

EXAM. 10. To find the greatest rectangle inscribed in, and the least isosceles triangle circumscribed about, a given semiellipse.

EXAM. 11. To determine the same for a given parabola. EXAM. 12. To determine the same for a given hyperbola. EXAM. 13. To inscribe the greatest cylinder in a given cone; or to cut the greatest cylinder out of a given cone. EXAM. 14. To determine the dimensions of a rectangular cistern, capable of containing a given quantity a of water, so as to be lined with lead at the least possible expense.

EXAM. 15. Required the dimensions of a cylindrical tan


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kard, to hold one quart of ale measure, that can be made of the least possible quantity of silver, of a given thickness. EXAM, 16. To cut the greatest parabola from a given cone. Exam. 17. To cut the greatest ellipse from a given cone. EXAM. 18. To find the value of x when x is a minimum.


92. The Method of Tangents, is a method of determining the quantity of the tangent and subtangent of any algebraic curve; the equation of the curve being given. Or, vice versa, the nature of the curve, from the tangent given.

If AE be any curve, and E be any point in it, to which it is required to draw a tangent TE. Draw the ordinate ED: then if we can determine the subtangent TD, limited between the ordinate and tangent, in the axis produced, by joining the points, T, E, the line TE will be the tangent sought.


93. Let dae be another ordinate, indefinitely near to DE, meeting the curve, or tangent produced in e; and let ɛe be parallel to the axis AD. Then is the elementary triangle Eea similar to the triangle TDE; and

ea dЕ ED: DT.

· ⱭE ・・ flux. ED : flux. ad. flux. ED: flux. AD:



That is

DE. dt.




which is therefore the general value of the subtangent sought;
where x is the absciss AD, and y the ordinate DE.
Hence we have this general rule.


94. By means of the given equation of the curve, when put into fluxions, find the value of either or y or of




which value substitute for it in the expression DT =



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and, when reduced to its simplest terms, it will be the value of the subtangent sought.


e equation Curve be that which is defined,

EXAM. 1. Let the er expressed by the equation ax+xy2 y3=0. Here the fluxion of the equation of the curve is 2axx + y2x+2xyy-3y3y=0; then, by transposition, 2αxx + y2x=3y3y-2xyy; and hence, by division, yx 3y3 - 2xy2

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3y2 — 2xy


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2ax + y2 which is the value of the subtangent TD sought.

EXAM. 2. To draw a tangent to a circle; the equation of which is ax-x=y2; where x is the absciss, y the ordinate, and a the diameter.

EXAM. 3. To draw a tangent to a parabola; its equation being ax = y; where a denotes the parameter of the axis. EXAM. 4. To draw a tangent to an ellipse; its equation being c2 (ax-x2)=a2y2; where and c are the two axes. EXAM. 5. To draw a tangent to an hyperbola; its equation being ca(ax+2)=ay; where a and c are the two axes. c2 EXAM. 6. To draw a tangent to the hyperbola referred to the asymptote as an axis; its equation being xy-a2; where a2 denotes the rectangle of the absciss and ordinate answering to the vertex of the curve.


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95. RECTIFICATION, is the finding the length of a curve line, or finding a right line equal to a proposed curve. By art 10 it appears, that the elementary triangle Eae, formed by the increments of the absciss, ordinate, and curve, is a right-angled triangle, of which the increment of the curve is the hypothenuse and therefore the square of the latter is equal to the sum of the squares of the two former; that is, Ee2=Eα2+αe2. Or, substituting, for the increments, their proportional fluxions, it is zz xx + yy, or z

2 ✓ x2+ y2 where z denotes any curve line AE, & its absciss AD, and y its ordinate DE. Hence this rule.



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