Question

If a and b are positive numbers, find the maximum value of f(x)=x^a(1-x)^b, 0<=x<=1

Answer 1

\[0\le x \le1\]

Answer 2

\[x^a(1-x)^b\]

Answer 3

@Easyaspi314

Answer 4

@Hero

Answer 5

@Zarkon

Answer 6

finding maximum denotes getting the 1st derivative of the function and equate to zero...

Answer 7

ok so the derivative would be ax^(a-1)(1-x)^b-b(1-x)^(b-1)(x^a)

Answer 8

isn't it + since that d (uv) = u dv + v du

Answer 9

yeah but with the chain rule you have to take the derivative of (1-x) which is -1 and multiply it to the exponent rule part so it becomes -b

Answer 10

oh yes i overlook (1-x) (-1) dx...

Answer 11

so what do i do now

Answer 12

combine like terms, arrange in d[f(x)]/dx then equate to 0 to solve for x in terms of a and b...

Answer 13

equating to zero (0) the 1st derivative denotes either maximum or minimum....

Answer 14

i have b(1-x)^b-1(x^a)=ax^(a-1)(1-x)^b

Answer 15

\[x^a(1-x)^b=>bx^a(1-x)^{b-1}+a(1-x)^b x^{a-1}=0\]
solve for x=?

Answer 16

it should be -

Answer 17

oh yes

Answer 18

\[bx^a(1-x)^{b-1}=a(1-x)^b x^{a-1}\]

Answer 19

oops.. we can cancel (1-x)^b on both sides...

Answer 20

exactly^ :D sorry im eating so didnt type..was waiting fr sm1 to observe that!

Answer 21

but its (1-x)^b-1 on the left

Answer 22

\[\frac{bx ^{a}}{(1-x)}=ax ^{a-1}\]

Answer 23

we can further cancel (x^a)...

Answer 24

\[\LARGE bx^a(1-x)^{b-1}=a(1-x)^b x^{a-1}\]
\[\frac{bx^a \cancel{(1-x)^{b}}}{(1-x)}=a \cancel{(1-x)^b} x^{a-1}\]

Answer 25

\[\frac{ b }{ 1-x }=\frac{ a }{ x }\]

Answer 26

solve for x=0..and solve for maxima minima

Answer 27

if x=0 then the left side of @Orion1213 equation is undefined

Answer 28

bx = a - ax
bx + ax = a
(a+b)x = a
x = a/(a+b)

Answer 29

did u get it @muzzammil.raza

Answer 30

i got lost after orion said b/1-x=a/x and then you said x=0 which makes the equation undefined

Answer 31

we have solve x here, the condition must be satisfied that a and b are both positive numbers...

Answer 32

and maximum value of f(x) occurred at x=a/(a+b)...

Answer 33

... regarding the condition for x.... 0<=x<=1... if a is positive, it should be greater than 0 so x is safe of not to be zero....

Answer 34

but how did you get from b/1-x=a/x to bx=a-ax

Answer 35

if we go back to...
\[\frac{ bx ^{a} }{ 1-x }=ax ^{a-1}\]

Answer 36

we got this form when we cancel (1-x)^b...

Answer 37

yes i get that part

Answer 38

examining further the equation, we can further cancel x^a...

Answer 39

note on the right-side...
\[x ^{a-1}=x ^{a}\times x ^{-1}=\frac{ x ^{a} }{ x }\]

Answer 40

so we can re-write the equation as...
\[\frac{ bx ^{a} }{ 1-x }=\frac{ ax ^{a} }{ x }\]

Answer 41

if we multiply both sides by 1/(x^a)... x^a will cancel and leaves the equation as
\[\frac{ b }{ 1-x }=\frac{ a }{ x }\]

Answer 42

from this form... we can now solve for the value of x... multiply both sides by x(1-x)....
\[x \cancel{(1-x)}\left[ \frac{ b }{ \cancel{(1-x)} } \right]=\left[ \frac{ a }{ \cancel{x} } \right](1-x)\cancel{x}\]

Answer 43

bx = a (1-x)
bx = a - ax
ax + bx = a
(a+b)x = a
\[x=\frac{ a }{ a+b }\]

Answer 44

@muzzammil.raza , is it ok now... got the idea?

Answer 45

so the maximum is a/a+b

Answer 46

not yet... the requirement is the maximum of f(x)... so we need to substitute x in our function... and that is the answer we are looking for....

Answer 47

\[f(x) = x ^{a}(1-x)^{b}\]

Answer 48

ok so the maximum is (a/(a+b))^a(1-(a/a+b))^b

Answer 49

yes... that is the one we're looking for... :)

Answer 50

can i simplify that

Answer 51

it seems complicated to me already... :)

Answer 52

is that a yes or no

Answer 53

for me... no...

Answer 54

ok i will leave it then thanks for all the help

Answer 55

no problem... :)