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anonymous
 5 years ago
If a and b are positive numbers, find the maximum value of f(x)=x^a*(1x)^b, 0 less than or equal to x less than or equal to 1.
Your answer may depend on a and b. What is the maximum value?
anonymous
 5 years ago
If a and b are positive numbers, find the maximum value of f(x)=x^a*(1x)^b, 0 less than or equal to x less than or equal to 1. Your answer may depend on a and b. What is the maximum value?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1maximum value of: \[f(x)=x^a*(1x)^b;\ 0\le x \le 1\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[f'(x)=ax^{(a1)}(1x)^bx^ab(1x)^{(b1)}\] \[ax^{(a1)}(1x)^bx^ab(1x)^{(b1)}=0\] solving this might get us some critical points

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Critical point = a/(a+b)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Got the answer. It is (a/(a+b))^a(b/(a+b))^b

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[ax^{(a1)}(1x)^b=x^ab(1x)^{(b1)}\] \[ax^{(a1)}x^{a}(1x)^b=b(1x)^{(b1)}\] \[ax^{1}(1x)^b=b(1x)^{(b1)}\] \[a=xb(1x)^{(b1)}(1x)^{b}\] \[a=xb(1x)^{1}\] \[\frac{a}{b}=\frac{x}{1x}\] \[aax=bx\] \[a=bx+ax\] \[\frac{a}{b+a}=x\] looks to be that way if i did it right
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