Question:
Let f(x) = min{2x2,52−5x}, where x is any positive real number. Then the maximum possible value of f(x) is
Correct Answer: 32
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f(x) = min (${2x^{2},52-5x}$)
The maximum possible value of this function will be attained when $2x^{2}=52-5x$.
$2x^2+5x-52=0$
$(2x+13)(x-4)=0$
=> $x=\frac{-13}{2}$ or $x = 4$
Since x has to be positive integer, we can discard the case $x=\frac{-13}{2}$.
$x=4$ is the point at which the function attains the maximum value.
putting $x=4$ in the original function, we get, $2x^2 = 2*4^2= 32$.
Or the maximum value of f(x) = $32$.
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