Given that xy =3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x + 3y/5. Please show steps. Thank you!

Question

Given that xy =3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x + 3y/5. Please show steps. Thank you!​

Sapo1 · Answer

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 $F(x,y)=10x+\frac{3y}{5}$

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

$F(x)=10x+\frac{9}{10x}$

==> F'(x) = $10-\frac{9}{10x^{2} }$

Setting this to 0 to find the critical point, we get $10=\frac{9}{10x^{2} }$

or $x^{2} = \frac{9}{100}$

==> x = 3/10. (Chose the positive because of the hypothesis.)

==> y = 5.

Now the 2nd derivative is F” = $\frac{9}{5x^{3} } > 0$

hen x = 3/10.

Hence an (absolute) minimum exists at (3/10,5), with value

$F(\frac{3}{10} , 5) = 10 *\frac{3}{10} +\frac{3*5}{5} =6$