Question 1026931
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 {{{F(x,y) = 10x + (3y)/5}}}.

Since {{{y = 3/(2x)}}}, ==> F(x,y) becomes, after substitution, 

{{{F(x) = 10x + 9/(10x)}}}

==> F'(x) = {{{10 - 9/(10x^2)}}}
Setting this to 0 to find the critical point, we get {{{10 = 9/(10x^2)}}}, or
{{{x^2 = 9/100}}} ==> x = 3/10.  (Chose the positive because of the hypothesis.)
==> y = 5.
Now the 2nd derivative is F" = {{{9/(5x^3) > 0}}} when x = 3/10.
Hence an (absolute) minimum exists at (3/10,5), with value
{{{F(3/10,5) = 10(3/10) + (3*5)/5 = 6}}}.