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Given that $xy = \dfrac32$ and both $x$ and $y$ are nonnegative real numbers, find the minimum value of $10x + \dfrac{3y}5.$

 Jul 3, 2019
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\(xy = \dfrac32 \)       \(10x + \dfrac{3y}5 \)

 

y = 3/[2x]

 

So....let f(x)  =   10x + 3 [ 3]/ [ 5 * 2x]   =  10x + 9/[10x ]  =  10x + (9/10)x^-1

 

Take the derivative , set to 0 and solve for x

 

10 -  (9/10)x^-2  =  0

 

10 = (9/10)x^-2

 

10x^2  = 9/10

 

x^2  = 9/100

 

x = 3/10

 

And we can prove that  x = 3/10 minimizes f(x)  by evaluating  f " (3/10)

 

2(9/10)

_______  >  0       so   x = 3/10   minimizes f(x)

(3/10)^3

 

And  y  = 3 /[2(3/10)]  =  3 / (6/10)  =  30/6  =  5

 

And   f(x)  =  10(3/10)  +  3(5)/5  =  3 + 3  = 6    = minimum for f(x)

 

 

cool cool cool

 Jul 3, 2019
edited by CPhill  Jul 3, 2019

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