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: $(-\infin, \infin)$

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

Find the critical number(s)

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

$$x^2+x-2=0$$

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

$$x_1=-2, x_2=1$$

Critical numbers: $-2, 1.$

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

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

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

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

$$f''(x)=0=>12x-6=0=>x=\dfrac{1}{2}$$

$$f(\dfrac{1}{2})=2(\dfrac{1}{2})^3+3(\dfrac{1}{2})^2-12(\dfrac{1}{2})+1=-4$$

If $x<\dfrac{1}{2}, f''(x)<0, f(x)$ is concave down.

If $x>\dfrac{1}{2}, f''(x)>0, f(x)$ is concave up.

Point $(\dfrac{1}{2}, -4)$ is an inflection point.

Point $(-2, 21)$ is a local maximum.

Point $(1, -6)$ is a local minimum.

Point $(\dfrac{1}{2}, -4)$ is an inflection point.

(b)

Let $x=$ the first number, let $y=$ the second number.

Given $x+y=S, x>0, y>0.$

Then $y=S-x, 0<x<S,$

The product of these numbers is $P=xy.$

Substitute

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

Find the first derivative with respect to $x$

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

Find the critical number(s)

$$P'(x)=0=>S-2x=0=>x=\dfrac{1}{2}S$$

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

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

The function $P$ has a local maximum at $x=\dfrac{1}{2}S.$

Since the function $P$ has the only extremum on $(0, S),$ then the function $P$ has the absolute maximum on $(0, S)$ at $x=\dfrac{1}{2}S.$

$$y=S-\dfrac{1}{2}S=\dfrac{1}{2}S$$

$$P=\dfrac{1}{2}S(\dfrac{1}{2}S)=\dfrac{1}{4}S^2.$$

The maximum value of the product is $\dfrac{1}{4}S^2.$