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

Thanks!

WhichWitchIsWhich
Oct 23, 2017

#1**+2 **

\(xy=\frac32\qquad\rightarrow\qquad y=\frac{3}{2x} \\~\\ 10x+\frac{3y}{5}\,=\,10x+\frac{3(\frac{3}{2x})}{5}\,=\,10x+\frac{9}{10x}\)

Let's say that this is equal to something, " n "

n = \(10x+\frac{9}{10x}\)

We want to know the minimum value of n .

There might be a better way to do this, but we can find it by looking at a graph.

https://www.desmos.com/calculator/eg7s3kqh1j

Since x can't be negative, the minumum value of n is 6 .

hectictar
Oct 23, 2017

#2**+2 **

Thanks, hectictar.......here's another way, but...it uses some Calculus

xy = 3/2 → y = 3 / [ 2x ] (1)

10x + 3y/5 (2)

Sub (1) into (2)

10x + 9 / [10x] take the derivative of this and set it to 0

10 - (9/10)x^(-2) = 0 multiply through by x^2

10x^2 - (9/10) = 0 add 9/10 to both sides

10x^2 = 9/10 multiply through by 10

100x^2 = 9 divide both sides by 100

x^2 = 9 / 100 take the positive root

x = 3/10

And y is found using xy = 3/2

(3/10) y = (3/2) multiply both sides by 10/3

y = 5

So....the minimum is 10(3/10) + 3*5 /5 = 3 + 3 = 6

CPhill
Oct 23, 2017