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

2 Answers
Jun 29, 2015

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]}

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(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)