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frez [133]
3 years ago
9

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]
Mathematics
1 answer:
mafiozo [28]3 years ago
5 0

9514 1404 393

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.

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Perimeter ratio = 4 : 9

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Step-by-step explanation:

Volume ratio = \frac{\text{volume of small prism}}{\text{Volume of large prism}}

                      = \frac{128}{1458}

                      = \frac{64}{729}

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Scale factor = \sqrt[3]{\frac{\text{volume of small prism}}{\text{Volume of large prism}}}

                    = \sqrt[3]{\frac{64}{729}}

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Perimeter ratio = Scale factor

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                         = 4 : 9

Surface area ratio = (Scale factor)²

                              = (\frac{4}{9})^{2}

                              = \frac{16}{81}

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