Question:
Let f(x)=max{5x, 52-2x2}, where x is any positive real number. Then the minimum possible value of f(x) is
Correct Answer: 20
Given x is positive real number. The minimum value of the maximum $\left\{5 x, 52-2 x^{2}\right\}$ will occur when both the graphs intersect. i.e., when $5 x=52-2 x^{2}$
$2 x^{2}+5 x-52=0$
$2 x^{2}+13 x-8 x-52=0$
$x(2 x+13)-4(2 x+13)=0$
$(x-4)(2 x+13)=0$
$x=4$ or $\frac{-13}{2}$
When $x=4, f(x)=20$
Given x is positive real number. The minimum value of the maximum $\left\{5 x, 52-2 x^{2}\right\}$ will occur when both the graphs intersect. i.e., when $5 x=52-2 x^{2}$
$2 x^{2}+5 x-52=0$
$2 x^{2}+13 x-8 x-52=0$
$x(2 x+13)-4(2 x+13)=0$
$(x-4)(2 x+13)=0$
$x=4$ or $\frac{-13}{2}$
When $x=4, f(x)=20$
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