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# Algebra help?

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

Thanks!

Oct 23, 2017

#1
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$$xy=\frac32\qquad\rightarrow\qquad y=\frac{3}{2x} \\~\\ 10x+\frac{3y}{5}\,=\,10x+\frac{3(\frac{3}{2x})}{5}\,=\,10x+\frac{9}{10x}$$

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 .

Oct 23, 2017
edited by hectictar  Oct 23, 2017
#2
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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   Oct 23, 2017