Find the maximum and minimum values of f(t)= `t*sqrt(64-t^2)` in the interval [-1,8]

Expert Answers

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The function f(t) = `t*sqrt(64 - t^2)`

f'(t) = `sqrt(64 - t^2) + t*(1/2)(1/sqrt(64 - t^2))*(-2t)`

=> `(64 - 2t^2)/sqrt(64 - t^2)`

Solving f'(t) = 0 gives `(64 - 2t^2)/sqrt(64 - t^2) = 0`

=> `64 - 2t^2 = 0`

=> `32 - t^2 = 0`

=>` t = sqrt 32`

=> `t = 4*sqrt 2`

At t =` 4*sqrt 2` , f''(t) is negative.

The function `f(t) = t*sqrt(64 - t^2)` has a maximum value at `t = 4*sqrt 2`

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