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Let x and y be nonnegative real numbers. If xy = \frac{2}{5}, then find the minimum value of 6x^2 + \frac{3}{5}y^2 - 2x + 8y.

 Mar 10, 2024
 #1
avatar+129473 
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\(6x^2 + \frac{3}{5}y^2 - 2x + 8y \)

xy = 2/5  →  y = 2 / (5x)

 

6x^2  + (3/5)(2 / (5x)^2 - 2x + 8(2 / (5x))

 

6x^2  + (12/125)x^(-2) - 2x + (16/5)x^(-1)

 

 

The graph shows that the minimum ≈ -26.525  when  x ≈ -.06

 

cool cool cool

 Mar 10, 2024

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