Question

If the sum of two positive numbers x and y is 3, then the absolute maximum of x^3+12xy is what?

Answer 1

Answer:

32 at x = 2

Step-by-step explanation:

If the sum of two positive numbers x and y is 3, then

$x+y=3\Rightarrow y=3-x$

and

$x^3+12xy=x^3+12x(3-x)=x^3-12x^2+36x$

To find the absolute maximum of the function

$f(x)=x^3-12x^2+36x$

find the derivative

$f'(x)=3x^2-24x+36$

and equate it to 0:

$3x^2-24x+36=0\\ \\x^2-8x+12=0\\ \\D=(-8)^2-4\cdot 12=64-48=16\\ \\x_{1,2}=\dfrac{-(-8)\pm \sqrt{16}}{2\cdot 1}=\dfrac{8\pm 4}{2}=6,\ 2$

For $x<2,\ f'(x)>0$ - the function f(x) increases

For $2<x<6,\ f'(x)<0$ - the function f(x) decreases

For $x>6,\ f'(x)>0$ - the function f(x) increases

So, x = 2 is maximum, x = 6 is minimum

The maximum value of $x^3+12xy$ is

$2^3+12\cdot 2\cdot (3-2)=8+24=32$