Question

$\sf \implies\: find\: the\: maximum\: and \: minimum\: values\: of \: function\: y\:=2x^{3}\:-15x^{2}\:+36x\:+1$

Answer 1

HELLO DEAR,



GIVEN:- f(x) = 2x³ - 15x² + 36x + 1

so, f'(x) = 6x² - 30x + 36

for Maxima/minima , f'(x) = 0

6x² - 30x + 36 = 0

=> x² - 5x + 6 = 0

=> x² - 3x - 2x + 6 = 0

=> x(x - 3) - 2(x - 3) = 0

=> (x - 2)(x - 3) = 0

=> x = 2 , x = 3

and, f''(x) = 12x - 30

f''(2) = 12(2) - 30 = -6 >0

therefore, x = 2 is a point of Maxima

and f''(3) = 12(3) - 30 = 6>0

therefore, x = 3 is a point of minima

hence, maximum value at x = 2 is f(2) = 16 - 60 + 72 + 1 = 29

and

minimum value at x = 3 is f(3) = 54 - 135 + 108 + 1 = 28


I HOPE IT'S HELP YOU DEAR,
THANKS

Answer 2

$<b>Given Expression is y = 2x^3 - 15x^2 + 36x + 1.</b>$

On differentiating, we get

⇒ y' = 6x^2 - 30x + 36

We need to set the derivative equal to 0 in order to find maximum and minima.  

⇒ y' = 0

⇒ 6x^2 - 30x + 36 = 0

⇒ 6(x^2 - 5x + 6) = 0

⇒ x^2 - 5x + 6 = 0

⇒ x^2 - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 2)(x - 3) = 0

⇒ x = 2,3..

$<b>Differentiate the given equation again with respect to x.</b>$

⇒ y'' = 12x - 30.

Now,

$<b>Substitute x = 2 in y'':</b>$

⇒ 12x - 30

⇒ 12(2) - 30

⇒ -6.

$<b>Substitute x = 3 in y'':</b>$

⇒ 12x - 30

⇒ 12(3) - 30

⇒ 36 - 30

⇒ 6.

Hence,

$<b>At x = 2, the maximum value is -6.</b>$

$<b>At x = 3, and the minimum value is 6.</b>$

Hope it helps!