Finding the maximum value given an interval

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To find the maximum value of the function f(x) = x^a(1 - x)^b on the interval 0 ≤ x ≤ 1, the first derivative f'(x) must be set to zero. The critical points include x = 0, x = 1, and another derived from the equation (a(1-x) + bx) = 0. It's essential to substitute these x-values back into the original function to determine the maximum value rather than just the x-coordinate. A common mistake was identified regarding a typo with a minus sign, which affected the calculations.
PandaherO
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Homework Statement



If a and b are positive numbers, find the maximum value of f(x) = x^a(1 - x)^b on the interval 0 ≤ x ≤ 1.?

Homework Equations



f'(x) = x^a . -b(1-x)^(b-1) + (1-x)^b . ax^(a-1)

The Attempt at a Solution



I know to set f'(x) = 0 but I'm not sure how I'm supposed to tidy up the f'(x) above..
 
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Either x or (1-x) are solutions, or otherwise you can divide away x^a and (1-x)^b.
 
PandaherO said:

Homework Statement



If a and b are positive numbers, find the maximum value of f(x) = x^a(1 - x)^b on the interval 0 ≤ x ≤ 1.?

Homework Equations



f'(x) = x^a . -b(1-x)^(b-1) + (1-x)^b . ax^(a-1)

Set f '(x) equal to zero.

xa b(1-x)(b-1) + (1-x)b ax(a-1) = 0

Now factor out (x(a-1) (1-x)(b-1)).

There is one solution in addition to the two mentioned by I like Serena.
 
the question also gave a hint that my maximum value may depend on a and b

so (x^(a-1) (1-x)^(b-1)) (a(1-x)+bx) =0

then do I set (a(1-x)+bx)=0?
and solve for x?

i got x=a/(a-b) and it isn't the correct answer...
 
You made a typo with a minus sign, giving you the wrong result.

Furthermore, if I read your problem correctly, it asks for the maximum value, not the x-coordinate of the maximum value.

You have 3 solutions: x=0, x=1, and this one.
You should substitute those and pick the greatest for the maximum value.
 
I like Serena said:
You made a typo with a minus sign, giving you the wrong result.

Furthermore, if I read your problem correctly, it asks for the maximum value, not the x-coordinate of the maximum value.

You have 3 solutions: x=0, x=1, and this one.
You should substitute those and pick the greatest for the maximum value.

OHH no wonder...*facepalms myself* the minus sign... and yea, the problem wanted the x and y coordinate)
Thank you so much owo
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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