What is the extreme 20(t/(12)-ln(t/(12)))+30 ?

The extreme 20(t/(12)-ln(t/(12)))+30 is Minimum(12,50)

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extreme 20(t/(12)-ln(t/(12)))+30

extreme points

20 (\frac{t}{12} - ln (\frac{t}{12})) + 30

Minimum (12, 50)

Solution steps

f^{'} (t) = - 20 (\frac{1}{t} - \frac{1}{12})

Find intervals:

Decreasing

: 0 < t < 12,

Increasing

: 12 < t < \infty

Plug

t = 12

into

20 (\frac{t}{12} - ln (\frac{t}{12})) + 30 : 50

Minimum (12, 50)

f (x) = cos (x) - 9 x

f (x, y) = e^{x^{2} + x y + y^{2}}

f (x) = - x^{2} + 10

f (x) = - x^{3} + 4 x^{2} + 3 x - 2

f (x) = \frac{x \cdot e ^{x}}{e ^{x} - e}

What is the extreme 20(t/(12)-ln(t/(12)))+30 ?
The extreme 20(t/(12)-ln(t/(12)))+30 is Minimum(12,50)