If y= 3x + 6, what is the minimum value of (x^3)(y)?

Question

If y= 3x + 6, what is the minimum value of (x^3)(y)?

Answer 1

x³y=x³(3x+6)=3x⁴+6x³

(3x⁴+6x³)'=12x³+18x²=0, x=0, x=-3/2

min(x³y)=-81/16, for x=-3/2.

 

 

Answer 2

F(x) = x^3*y = x^3 * (3x+6)  = 3x^4 + 6x^3

 

the derivative is F'(x) =  12x^3 + 18x^2

 

F'(x) = 0 = 12x^3 + 18x^2

 

               = 6x^2 ( 2x + 3)

 

   Critical values at x=0 and x = -3/2

 

Critical points (0,0)

       

[3(-3/2)+6] * (-3/2)^3 = [-9/2 + 6] *(-27/8)

                                   = 3/2*(-27/8) 

                                  = -81/16

                                 

 

(-3/2, -81/16)