Answer to Question #212390 in Calculus for Kenneth

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

(a)

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

Domain: $(-\infin, \infin)$ as polynomial.

Find the first derivative with respect to $x$

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

Find the critical number(s)

$$f'(x)=0=>6x^2+6x-12=0$$

$$6(x-1)(x+2)=0$$

Critical numbers: $-2, 1.$

By the First Derivative Test

If $x<-2,$ then $f'(x)>0, f(x)$ increases.

If $-2<x<1,$ then $f'(x)<0, f(x)$ decreases.

If $x>1,$ then $f'(x)>0, f(x)$ increases.

$$f(-2)=2(-2)^3 + 3(-2)^2 - 12 (-2) +1=21$$

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

The function $f$ has a local maximum with value of $21$ at $x=-2.$

The function $f$ has a local minimum with value of $-6$ at $x=1.$

(b) Let $x=$ the first number, $x>0.$ Then the second number will be $S-x.$

Given $S-x>0.$

Their product will be

$$f(x)=x(S-x), 0<x<S$$

Find the first derivative with respect to $x$

$$f'(x)=( x(S-x))'=S-2x$$

Find the critical number(s)

$$f'(x)=0=>S-2x=0$$

$$x=\dfrac{1}{2}S$$

Critical number: $\dfrac{1}{2}S .$

By the First Derivative Test

If $x<\dfrac{1}{2}S,$ then $f'(x)>0, f(x)$ increases.

If $x>\dfrac{1}{2}S,$ then $f'(x)<0, f(x)$ decreases.

$$f(\dfrac{1}{2}S)=\dfrac{1}{2}S(S-\dfrac{1}{2}S)=\dfrac{1}{4}S^2$$

The function $f$ has a local maximum with value of $\dfrac{1}{4}S^2$ at $x=\dfrac{1}{2}S.$

Since the function $f$ has the only extremum, then the function $f$ has the absolute maximum with value of $\dfrac{1}{4}S^2$ at $x=\dfrac{1}{2}S$ for $0<x<S.$

The maximum value of the product $\dfrac{1}{4}S^2$ will be if we take two equal positive numbers

$$I\ number=\dfrac{1}{2}S=II\ number$$