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Let x and y be nonnegative real numbers. If xy = \frac{2}{5}, then find the minimum value of 6x + \frac{3}{5y}.

 Apr 4, 2024
 #1
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6x + (3/5)y

xy = 2/5

y = 2 /(5x)

 

6x + (3/5)(2) / (5x)

 

6x  + (6/25)x^(-1)           take the derivative and set to 0

 

6 - (6/25)x^(-2)  =  0    multiply

 

6x^2  = (6/25)

 

x^2 =  1/25

 

x =   1/5

 

Take the scond derivative to see that this is a  min

 

 2(6/25)(.2)^(-3)  =  60   which is > 0 ....so this is a  min

 

Min  at (.2, 2)    =  6(.2) + (3/5)(2)  =  1.2 + 1.2  =  2.4

 

cool cool cool

 Apr 5, 2024

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