The maximum value of the function $f (x) = 2 x^{3} - 15 x^{2} + 36 x - 48$ on the set $A = {x | x^{2} + 20 \leq 9 x}$ is

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Solution

$x ² + 20 \leq 9 x$

$x ² - 9 x + 20 \leq 0$

$x ² - 4 x - 5 x + 20 \leq 0$

$x (x - 4) - 5 (x - 4) \leq 0$

$(x - 4) (x - 5) \leq 0$

$4 \leq x \leq 5$

$A = {4 \leq x \leq 5}$

now,
$f (x) = 2 x ³ - 15 x ² + 36 x - 48$

differentiate wrt $x$

$f^{'} (x) = 6 x ² - 30 x + 36 = 6 (x ² - 5 x + 6)$

$= 6 (x - 2) (x - 3)$

So $f (x)$ is increasing in $(- \infty, 2) \cup (3, \infty)$

maximum value of $f (x)$ at $x = 5$

$f (5) = 2 (5) ³ - 15 (5) ² + 36 (5) - 48$

$= 250 - 375 + 180 - 48$

$= 430 - 423 = 7$

$∴ f_{max} = f (5) = 7$

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