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Difference between revisions of "1962 AHSME Problems/Problem 26"

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==Solution==
 
==Solution==
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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>
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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{</math>\text{E}\ \dfrac{16}{3}}$

Revision as of 20:43, 22 April 2018

Problem

For any real value of $x$ the maximum value of $8x - 3x^2$ is:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{8}3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \frac{16}{3}$

Solution

Let $f(x) = 8x-3x^2$ Since $f(x)$ is a quadratic and the quadratic term is negative, the maximum will be $f(- \dfrac{b}{2a})$ when written in the form $ax^2+bx+c$. We see that $a=-3$, and so $- \dfrac{b}{2a} = -( \dfrac{8}{-6}) = \dfrac{4}{3}$. Plugging in this value yields $f(\dfrac{4}{3}) = \dfrac{32}{3}-3 \cdot \dfrac{16}{9} = \dfrac{32}{3} - \dfrac{16}{3} = \boxed{$ (Error compiling LaTeX. Unknown error_msg)\text{E}\ \dfrac{16}{3}}$

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