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Let there be two numbers a and b such that their sum is n, n > 0. What numbers should you pick in order to maximize the number a*b?
Answer:
a = a
a + b = n <=> b = n - a
=>
a*b = a(n - a) = na - a^2 = f(a)
f'(a) = 0
n - 2a = 0
2a = n
a = n/2
n/3 < n/2
f'(n/3) = n - (2n)/3 = n/3 > 0
2n/3 > n/2
f'(2n/3) = n - (4n)/3 = -n/3 < 0
=> f(n/2) is the maximum
b = n - a = n - n/2 = n/2
=> maximum number a*b = (n^2)/4
Let there be two positive numbers a and b. Show that their geometric mean is at most their arithmetic mean.
Answer:
(ab)^(1/2) <= (a + b)/2
ab <= (a^2 + 2ab + b^2)/4
a^2/4 + ab/2 - ab + b^2/4 >= 0
a^2/4 - ab/2 + b^2/4 >= 0
a^2 - 2ab + b^2 >= 0
(a - b)^2 >= 0
True for all positive numbers a and b. The equality happens when a = b.
Thus,
(ab)^(1/2) <= (a + b)/2
I am Jesse Sakari Hyttinen and I will see you in the next post!
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