Extreme problem help needed

Question

Consider the function B(x)=x^a(1-x)^b on 0≤x≤1, with a,b>0.

Absolute Maximum value: ______

Answer 1

Hi Odai,

The function B(x)=x^a (1-x)^b on [O,1]

a) continuous

b) is not constant but assumes same values at both ends of the interval [0,1]: B(0)=B(1)=0.

c) For x between 0 and 1 B(x)>0.

These properties together indicate that the function must reach at least one maximum somewhere between 0 and 1

At points x_m where B(x) reaches an extremum (maximum or minimum)

dB/dx=0 (1)

Using the rules

d [f(x) g(x)]/dx={[df(x)/dx]g(x) + f(x)[dg(x)/dx]} (2)

and for any constant a

d[f(x)^a]=af(x)^(a-1) (3)

we would get

[ax^(a-1)](1-x)^b- (x^a)b[(1-x)^(b-1)]=0

or ax^(a-1)](1-x)^b-=(x^a)b[(1-x)^(b-1)] (4)

There are 3 solutions of (3):

x=O; x=1 and ty he third solution we'd get by

multiplying and dividing the 1st term in l.h.s of (4) by x and

the 2nd term - by by (1-x) :

(a/x) B(x) - b/(1-x) B(x)=[a/x- b/(1-x)]B(x)=0 (5)

Because B(x)≠0 for 0<x<1, (5) is equivalent to

a/x - b/(1-x)=0 (6)

After adding fractions in (6) we would get an equation for x:

[a(1-x)-bx]/[x(1-x)]=0 ⇒ (a+b)x=a with a (unique) solution

x=a/(a-b) [from which (1-x)=b/(a+b)] (7)

So a/a+b is the only point at which B(x) reaches a nonzero (positive) extremum

(as was noted above, that must be the maximum x_m

(Substituting x_m and (1- x_m) back into B(x) we get the maximum value of B(x)

[B(x)]_max=B(x_m)={[a/(a+b)]^a}{[b/(a+b)]^b}=

(a^a b^b)/(a+b)^(a+b) (8)

A note: there is a way to directly prove that the extremum at x= a/a+b is indeed a maximum.,Another words, let's prove that the 2nd derivative of B(x) at x=a/a+b is negative.

For that, let's take the 1st derivative of B(x) [l.h.s of (5)]

and make a derivative of it:

d²B(x)/dx²=d{[a/x- b/(1-x)]B(x)}/dx=

[-(a/x²)-(b/(1-x)²]B(x)+[a/x-b/(1-x)][dB(x)/dx] (9)

Because we are looking at the value of d²B(x)/dx² at the point of extremum x =a/(a+b), the 2nd term in (9) is zero b/c at a point of extremum dB(x)/dx=0

As to the 1st term of (9) it must be negative b/c B(x) is positive in the interval (0,1). So, negativity of the 2nd derivative at the extremal point a/a+b is proven; therefore at that point the function B indeed reaches it's (absolute) maximum.

Hope it is helpful

Dr.Ariel B.

Answer 2

B(x) has a absolute maximum value for only a = b = 1.

https://www.desmos.com/calculator/w8zwbxbcjr

Answer 3

B(x) = x^a(1-x)^b

B'(x) = ax^a-1(1-x)^b - b(1-x)^b-1x^a = x^a-1(1-x)^b-1[a(1 - x) - bx] = x^a-1(1-x)^b-1[a - x(a + b)] = 0

x = 0, 1, a/(a+b)

B(0) = B(1) = 0

B(a / (a+b)) = [a^a / (a+b)^a][1 - (a / (a+b))]^b = [a / (a+b)]^a [b / (a+b)]^b = a^ab^b / (a + b)^a+b > 0

Absolute max value = a^ab^b / (a + b)^a+b

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