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

2 Answers
Jun 29, 2015

Answer:

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

Answer:

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#)