Answer to Question #213699 in Calculus for amit

Question #213699

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.



1
Expert's answer
2021-08-03T16:07:23-0400

Solution:

(a):

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

Put "f'(x)=0"

"6x^2+6x-12=0\n\\\\\\Rightarrow x^2+x-2=0\n\\\\\\Rightarrow x^2+2x-x-2=0\n\\\\\\Rightarrow (x+2)(x-1)=0\n\\\\\\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\n\\\\\\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\n\\\\\\Rightarrow \\dfrac{dP}{dy}=S-2y\n\\\\ Put\\ \\dfrac{dP}{dy}=0\n\\\\\\Rightarrow S-2y=0\n\\\\\\Rightarrow S=2y\n\\\\\\Rightarrow x+y=2y\n\\\\\\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"



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