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For what real value of b is the expression \(\frac{1}{2}b^2 + 5b - 3\)minimized?

 Apr 28, 2020
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One way to do this is to find the vertex of this quadratic:  y  =  0.5b2 + 5b - 3

 

Move the 3 to the other side:                                      y + 3  =  0.5b2 + 5b

Factor out the 0.5:                                                       y + 3  =  0.5(b2 + 10b)

Complete the square:                                       y + 3 + 12.5  =  0.5(b2 + 10b + 25)

Factor and simplify:                                                 y + 15.5  =  0.5(b + 5)2

 

The vertex is (-5, -15.5) which means that the minimum is -15.5 and this occurs at -5.

 Apr 28, 2020

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