SOLUTION: Given that xy = 3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x + (3y)/(5).

Question 1026931: Given that xy = 3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x + (3y)/(5).
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
First of all, x and y cannot be zero because of the condition xy = 3/2. Hence x and y can only be positive.
Now let .
Since , ==> F(x,y) becomes, after substitution,

==> F'(x) =
Setting this to 0 to find the critical point, we get , or
==> x = 3/10. (Chose the positive because of the hypothesis.)
==> y = 5.
Now the 2nd derivative is F" = when x = 3/10.
Hence an (absolute) minimum exists at (3/10,5), with value
.
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