What are the absolute extrema of #f(x)=x^(1/3)*(20-x)in[0,20]#?

1 Answer
Feb 14, 2018

Answer:

The absolute minimum is #0#, which occurs at #x = 0# and #x=20#.

The absolute maximum is #15root(3)5#, which occurs at #x = 5#.

Explanation:

The possible points that could be absolute extrema are:

  1. Turning points; i.e. points where #dy/dx = 0#

  2. The endpoints of the interval

We already have our endpoints (#0# and #20#), so let's find our turning points:

#f'(x) = 0#

#d/dx(x^(1/3)(20-x)) = 0#

#1/3x^(-2/3)(20-x) - x^(1/3) = 0#

#(20-x)/(3x^(2/3)) = x^(1/3)#

#(20-x)/(3x) = 1#

#20-x = 3x#

#20 = 4x#

#5 = x#

So there is a turning point where #x = 5#. This means that the 3 possible points that could be extrema are:

#x = 0" "" "x=5" "" "x=20#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let's plug these values into #f(x)#:

#f(0) = (0)^(1/3) (20 - 0) = 0 * 20 = color(red)0#

#f(5) = (5)^(1/3) (20 - 5) = root(3)(5) * 15 = color(red)(15root(3)5#

#f(20) = (20)^(1/3) (20-20) = root(3)(20) * 0 = color(red)0#

Therefore, on the interval #x in [0, 20]#:

The absolute minimum is #color(red)0#, which occurs at #x = 0# and #x=20#.

The absolute maximum is #color(red)(15root(3)5)#, which occurs at #x = 5#.

Final Answer