We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
226
1
avatar

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
 #1
avatar+104025 
+2

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

 

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

11 Online Users