Given that 
and both 
and 
are nonnegative real numbers, find the minimum value of 
+23245 If xy = 3/2, then y = (3/2)x^(-1)
If z = 10x + (3/5)y, then z = 10x + (3/5)(3/2)x^(-1)
If a minimum exists, it will exist when z' = 0: z' = 10 -(9/10)x^(-2)
Set z' = 0: 0 = 10 -(9/10)x^(-2)
-10 = -(9/10)x^(-2)
100x^2 = 9
x^2 = 9/100
x = 3/10
+128566 xy = 3/2 solving for y, we have y =3/(2x) and substituting this into the other function for y, we have
10x + [3(3/(2x))/5] = 10x + 9/(10x) = 10x + (9/10)x^-1
And taking the derivative of this and setting it to 0, we have
10 - (9/10)x^-2 = 0
(9/10)x^-2 = 10
(9/10) = 10x^2
x^2 = 9/100 so... x = ±(3/10)
Take the second derivative and see if either of the critical points "plugged in" to it produce a positive result...a positive result indicates a minimum.....so we have
10 - (9/10)x^-2 both points plugged into this will = 0
This doesn't tell us anything
Pick two points on either side of -3/10 and see what their slopes are choosing -1 and -1/2, we have
10 - (9/10)(-1)^-2 = + and 10 - (9/10)(-1/2)^-2 = - so -3/10 is a relative max
Now, pick two points on either side of 3/10 and calculate the slopes choose (1/5) and 1
10 - (9/10)(1/5)^-2 = - and 10 - (9/10)(1)^-2 = + so x = 3/10 is a relative minimum
But, this function is continuous at everywhere but 0. And at x = 3/10, the fuction has a relative minimum value of 6.And at -3/10, the function has a relative maximum value of -6. And since the function has a relative max at this point, then no absolute minimum exists
See the graph here......https://www.desmos.com/calculator/sj5mqtcm9d
