a,b,c,d ∈ R
a.b.c.d=64
what is the min. value of 16/a + 1 /3b + 3/5c + 5/4d
second one
a,b ∈ R
a ²+ 4ab + 2b²
a.b max. whole number value
Solve the problem
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Back to your question
I assume that the denominators are a, 3b, 5c, and 4d.
So we have 16/a + 1/3b + 3/5c + 5/4(64/abc) = 16/a + 1/3b + 3/5c + 5abc/256
df/da = -16/a2 + 5bc/256
df/db = -1/3b2 + 5ac/256
df/dc = -3/5c2 + 5ab/256
We want all of these to be 0. Get a system of 3 equations in 3 unknowns.
16/a2 = 5bc/256
a2bc = 4096/5