Answer to Question #221892 in Calculus for Anuj

Question

a) find and classify the critical points of the functions f(x) = 2x^3 + 3x^2 - 12 x +1 into maximum, minimum and inflection points as appreciate.

(b) The sum of two positive numbers is S. find the maximum value of their product.

Answer 1

Solution:

(a):

$f(x) = 2x^3 + 3x^2 - 12 x +1 \\f'(x)=6x^2+6x-12$

Put $f'(x)=0$

$6x^2+6x-12=0 \\\Rightarrow x^2+x-2=0 \\\Rightarrow x^2+2x-x-2=0 \\\Rightarrow (x+2)(x-1)=0 \\\Rightarrow x=-2,x=1$

Critical points are $x=-2,x=1$

Now, $f''(x)=12x+6$

$f''(-2)=12(-2)+6=-18<0\Rightarrow \ Maxima$

$f''(1)=12(1)+6=18>0\Rightarrow \ Minima$

Thus, maximum value is $f(-2) = 2(-2)^3 + 3(-2)^2 - 12 (-2) +1=21$

And minimum value is $f(1) = 2(1)^3 + 3(1)^2 - 12 (1) +1=-6$

Now, put $f''(x)=0$

$12x+6=0 \\\Rightarrow x=-6/12=-1/2$ , this is point of inflection.

(b):

Let the first and second numbers be $x,y$ respectively.

$x+y=S$ (where S is constant)

$\Rightarrow x=S-y\ ...(i)$

Let their product be P.

Then, $P=xy=(S-y)y$ [Using (i)]

$P=Sy-y^2 \\\Rightarrow \dfrac{dP}{dy}=S-2y \\ Put\ \dfrac{dP}{dy}=0 \\\Rightarrow S-2y=0 \\\Rightarrow S=2y \\\Rightarrow x+y=2y \\\Rightarrow x=y$

Again differentiating $\dfrac{dP}{dy}$ :

$\Rightarrow \dfrac{d^2P}{dy^2}=0-2=-2<0\Rightarrow\ Maxima$

Thus, two numbers are equal.

So, $S=x+y=x+x=2x$

$\Rightarrow x=y=\dfrac S2$

Now, $P=xy=\dfrac S2.\dfrac S2=\dfrac {S^2}4$