1962 AHSME Problems/Problem 26: Difference between revisions
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{{ | Let <math>f(x) = 8x-3x^2</math> Since <math>f(x)</math> is a quadratic and the quadratic term is negative, the maximum will be <math>f(- \dfrac{b}{2a})</math> when written in the form <math>ax^2+bx+c</math>. We see that <math>a=-3</math>, and so <math>- \dfrac{b}{2a} = -( \dfrac{8}{-6}) = \dfrac{4}{3}</math>. Plugging in this value, we see that <math>f(\dfrac{4}{3}) = \boxed{\dfrac{16}{3}}</math> | ||
Revision as of 20:41, 22 April 2018
Problem
For any real value of 
the maximum value of 
is:
Solution
Let 
Since 
is a quadratic and the quadratic term is negative, the maximum will be 
when written in the form 
. We see that 
, and so 
. Plugging in this value, we see that 
