Question

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

Answer 1

Solving `V'(x)=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'(x)=12x^2-104x+160=4(3x^2-26x+40)`

3. Now we can find the global extreme values solving `V'(x)=0`:
   `3x^2-26x+40=0`
   `x_1=2` and `x_2=20/3`

4. We can also determine the nature of these two points by calculating the sign of the derivative:
   `V(x)<0` if `x<20/3 and x>2`
   `V(x)>=0` if `x>=20/3 or x<=2`
   So the function `V(x)` grows until `x=2` then it decease until `x=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]}

Answer 2

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(x) = x(10-2x)(16-2x)` on `[0,5]`

The function can be further factored, to get:

`V(x) = x(10-2x)(16-2x)`

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

`= 4x(x-5)(x-8)`

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

`V(x) = 4x(x^2-13x+40) = 4(x^3-13x^2+40x)`

So,
`V'(x) = 4(3x^2-26x+40) = 4(3x-20)(x-2)`

The critical numbers (in `(-oo, oo)`) are `20/3` and `2`

But `20/3 > 15/3 = 5`, so it is not in the interval `[0, 5]`
. `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(x) = 4x(x-5)(x-8)` so

`V(0) = 0`

`V(2) = 8(-3)(-6) = 144`

`V(5) = 0` (The middle factor `x-5`, is `0`)

Answer:

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