Difference between revisions of "1962 AHSME Problems/Problem 26"
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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 yields <math>f(\dfrac{4}{3}) = \dfrac{32}{3}-3 \cdot \dfrac{16}{9} = \dfrac{32}{3} - \dfrac{16}{3} = \boxed{ | + | 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\bigg(- \dfrac{b}{2a}\bigg)</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} = -\bigg( \dfrac{8}{-6}\bigg) = \dfrac{4}{3}</math>. Plugging in this value yields <math>f(\dfrac{4}{3}) = \dfrac{32}{3}-3 \cdot \dfrac{16}{9} = \dfrac{32}{3} - \dfrac{16}{3} = \boxed{\text{E}\ \dfrac{16}{3}}</math> |
Latest revision as of 21:43, 2 May 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 yields 
