Question

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0, 20]

Answer 1

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Answer:

• maximum: 15∛5 ≈ 25.6496392002
• minimum: 0

Step-by-step explanation:

The minimum will be found at the ends of the interval, where f(t) = 0.

The maximum is found in the middle of the interval, where f'(t) = 0.

  $f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)$

This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...

  f(5) = (∛5)(20-5) = 15∛5

The absolute maximum on the interval is 15∛5 at t=5.