Question

If y= 3x + 6, what is the minimum value of (x^3)(y)?

Answer 1

Answer:

Given the statement: if y =3x+6.

Find the minimum value of $(x^3)(y)$

Let f(x) = $(x^3)(y)$

Substitute the value of y ;

$f(x)=(x^3)(3x+6)$

Distribute the terms;

$f(x)= 3x^4 + 6x^3$

The derivative value of f(x) with respect to x.

$\frac{df}{dx} =\frac{d}{dx}(3x^4+6x^3)$

Using $\frac{d}{dx}(x^n) = nx^{n-1}$

we have;

$\frac{df}{dx} =(12x^3+18x^2)$

Set $\frac{df}{dx} = 0$

then;

$(12x^3+18x^2) =0$

$6x^2(2x + 3) = 0$

By zero product property;

$6x^2=0$   and 2x + 3 = 0

⇒ x=0 and x = $-\frac{3}{2} = -1.5$

then;

at x = 0

f(0) = 0

and  

x = -1.5

$f(-1.5) = 3(-1.5)^4 + 6(-1.5)^3 = 15.1875-20.25 = -5.0625$

Hence the minimum value of $(x^3)(y)$ is, -5.0625