x is

Suppose that increases beyond x = a; then the factor a negative, and the product x(ax) will be negative and will increase in absolute value; therefore the relative value of y diminishes. The process is similar if x takes negative values.

This problem has the following geometric interpretation: ax may be considered the base and x the altitude of a rectangle ABCD; then y will be the area of ABCD.

A

The condition that x + (ax) or AB + BC be
a constant a will here mean that the perimeter
of the rectangle is always a constant quantity
2a. Therefore, the problem may be stated B
geometrically as follows: Study the variation

a-x

FIGURE 2

of the area of a rectangle whose perimeter is a constant 2a.

a2

a

D

C

The variation of the area of ABCD, as x increases from 0 to a can be followed easily: when x = 0, the base is a and the altitude 0; therefore the area is 0. Suppose that the altitude increases from 0 to; then the area increases continually from 0 to the maximum value As a increases from to a, then the base decreases from a to 0, and, therefore, the area diminishes from the maximum value to that of a straight line of length a, and is zero. In this discussion it has been assumed that x can be neither greater than a nor negative.

2

4

a2

461. PROBLEM II.-Find the largest rectangle which can be inscribed in a given triangle.

Let a be the altitude and b the base of the given triangle ABC, and let x and y be the altitude and the base of the rectangle to be determined. The area of the rectangle DEFG

will be

A

D

M

E

a

T

But

Hence,

=

DE MA

b

a

= (ax-x)·

bx2- abx+au = 0.

x =

ab± Va2b2 - 4 abu

FIGURE 3

The largest value which u can have, for which the values of x are real, is that value of u which will make the radical zero, namely,

Therefore, the maximum rectangle which can be inscribed in a given triangle has one-half the area of the triangle.

462. Graph of variation of u =

9 and a 3, then it is necessary to consider

X'

+5

U'

B

FIGURE 4

X

Hence, as x increases from 0 to 3, u increases

27

from 0 to its maximum value then decreases

4

x ap

to 0. If x = 4, U = 12. For values of x greater than 3, u is negative, and, as proaches∞, u approaches. If x = −1, u = 12, therefore, as x approaches

600

u also approaches oo. The graph exhibiting

these results is shown by the accompanying figure, in which BP shows the maximum value of u; u =

27
4

463. PROBLEM III.-How does the area of a triangle vary if the perimeter and the base are constant?

Call 2s the perimeter, and a, b, c, the sides of the triangle, then it is known that its area is

By hypothesis two factors, s and sa, may be regarded as constant,

and the other two factors, s b and s

28

Put

c, as variables whose sum

c = a, a constant.

Hence, it would be necessary to consider the variation of

Here it will be necessary to consider the variation of the product x (a x), as in Problem I, and when the maximum value of this product has been found, the maximum value of y will be its square root times M. Since, by Problem I, the product x (ax) is a maximum when x = a - x, i. e., when x = y is also a maximum

a

2

, and this maximum value will be M√

464. The graph of the curve y = MV x (ax).

Suppose that M2= 4 and a =

a a
2

5; then y = 21 x (5 — x).

M.

On plotting the points P, P, P, and all intermediate points,

the graph of y = 21x (5- c) is the curve P1 PÅ PÅ

If

x5, then x(5x) is negative and the corresponding values of y are imaginary, and no part of the graph will lie to the right of the vertical line through P,. If, moreover, x<0, then x (5 -- x) is also negative and no part of the graph lies to the left of the vertical line through P. Had the sign of the radical been then the graph of y = 2Vx (5 x) would be the curve PP' P,. It is clear

from the figure that the maximum value of the area of the triangle when M4 and a = 5 is

465. PROBLEM IV.-Given that the sum of two numbers is constant; investigate the question, what is the graph of the sum of their squares.

Let a be the sum of the numbers, of which x is one and a the other, and the sum of whose squares is y; hence,

y = x2 + (a — x)2.

This expression may be written

or

: 2 (x2 — ax + 1 ) = 2 ( x2 — ax + 2a2 + «2),

= 2(x2—

y = 2 x2 · 2 ax + a2 = 2( x2

i. e..

a2

4

2

x

This equation shows that y is the sum of two positive quantities, one of which, a, is fixed, and the other, 2 (-) is a variable. The quantity y will take its smallest value, when(x) is zero, i. e., when x = If it is considered that < x

2'

is negative; but (x) will be a positive quantity, which must be added to to find the corresponding y, which will be greater than

a2

2

a2

2

the part found for y when x Similarly, if it is considered that

then (c) is positive, and the corresponding value of Therefore, by definition, y is a minimum

a2

2

The geometrical interpretation of the preceding problem, 4, will now be given.

On plotting the points P1, P, ... Poo
P, P. Poo and all intermediate
points, the corresponding graph (Fig. 6) is x'
obtained.

+5

As shown in the figure, the values of y decrease continuously from+∞ to 26, from 26 to 8, then increase from 8 to 26, from 26 too as x increases respectively from +2, from 2 to +5, from +5 to +∞o.

- to -1, from

- 1 to

467. PROBLEM V.-As was stated in the beginning of this chapter, the intention was to discuss problems of maxima and minima whose solutions could be reduced to the solution of a quadratic equation. The problems thus far considered are all special cases of the more general problem: When has the quadratic expression ax + bx + c a maximum or a minimum value?