# How do you find the global extreme values for V(x) = x(10 - 2x)( 16 - 2x) on [0,5]?

Jun 29, 2015

Solving $V ' \left(x\right) = 0$.

#### Explanation:

1. We put the function in the polynomial form because it's easier to derivate it: V(x)=x(10−2x)(16−2x)=(10x-2x^2)(16-2x)=4x^3-32x^2-20x^2+160x=4x^3-52x^2+160x

2. We calculate the first derivative of the function: $V ' \left(x\right) = 12 {x}^{2} - 104 x + 160 = 4 \left(3 {x}^{2} - 26 x + 40\right)$

3. Now we can find the global extreme values solving $V ' \left(x\right) = 0$:
$3 {x}^{2} - 26 x + 40 = 0$
${x}_{1} = 2$ and ${x}_{2} = \frac{20}{3}$

4. We can also determine the nature of these two points by calculating the sign of the derivative:
$V \left(x\right) < 0$ if $x < \frac{20}{3} \mathmr{and} x > 2$
$V \left(x\right) \ge 0$ if $x \ge \frac{20}{3} \mathmr{and} x \le 2$
So the function $V \left(x\right)$ grows until $x = 2$ then it decease until $x = \frac{20}{3}$ and then it grows again. So ${x}_{1}$ is a relative maximum and ${x}_{2}$ is a relative minimum.

5. We can verify it by seeing the graph of this function:

graph{4x^3-52x^2+160x [-3, 14, -200, 200]}

Jun 29, 2015

This is a closed interval type problem. The global extrema occur at critical numbers in the interval or at endpoints of the interval.

#### Explanation:

$V \left(x\right) = x \left(10 - 2 x\right) \left(16 - 2 x\right)$ on $\left[0 , 5\right]$

The function can be further factored, to get:

$V \left(x\right) = x \left(10 - 2 x\right) \left(16 - 2 x\right)$

 = x[(-2)(x-5)]9(-2)(x-8)]

$= 4 x \left(x - 5\right) \left(x - 8\right)$

We need the critical numbers, so we need the derivative of $V \left(x\right)$. We could use the three factor product rule ($\left(u v w\right) ' = u ' v w + u v ' w + u v w '$), but I think it will be easier to multiply the variable factors and differentiate the resulting cubic function.

$V \left(x\right) = 4 x \left({x}^{2} - 13 x + 40\right) = 4 \left({x}^{3} - 13 {x}^{2} + 40 x\right)$

So,
$V ' \left(x\right) = 4 \left(3 {x}^{2} - 26 x + 40\right) = 4 \left(3 x - 20\right) \left(x - 2\right)$

The critical numbers (in $\left(- \infty , \infty\right)$) are $\frac{20}{3}$ and $2$

But $\frac{20}{3} > \frac{15}{3} = 5$, so it is not in the interval $\left[0 , 5\right]$
. $2$ is in the interval.

The extrema occur at either the endpoints, ($0$ and $5$) or at the critical number ($x = 2$), so we just check the values of $V$ for those 3 values of $x$:

$V \left(x\right) = 4 x \left(x - 5\right) \left(x - 8\right)$ so

$V \left(0\right) = 0$

$V \left(2\right) = 8 \left(- 3\right) \left(- 6\right) = 144$

$V \left(5\right) = 0$ (The middle factor $x - 5$, is $0$)

The maximum value is $144$ (it occurs at $2$)
The minimum is $0$ (it occurs at both $0$ and $5$)