How do you find the critical numbers for $g (t) = t \sqrt{4 - t}$ $g(t) = t sqrt(4-t)$ to determine the maximum and minimum?

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Answer 1 · 2018-08-03T08:01:50

$P M A X (\frac{8}{3}; \frac{16 \cdot \sqrt{3}}{9})$ $PMAX(8/3;(16*sqrt(3))/9)$

Given $g (t) = t \sqrt{4 - t}$ $g(t)=tsqrt(4-t)$
we get by the product and the chain rule

$g'(t)=sqrt(4-t)+t*(1/2)(4-t)^(-1/2)*(-1)$
which simplifies to

$g'(t)=sqrt(4-t)-t/(2sqrt(4-t))$
so we get

$g*(t)=0$ if

$sqrt(4-t)=t/(2sqrt(4-t))$ Multiplying with $2sqrt(4-t)$

$2(4-t)=t$

so $t=8/3$

$g''(t)=1/2(4-t)^(-1/2) * (-1)-1/2*(4-t)^(-1/2)+t/2(4-t)^(-3/2) * (-1/2)$

$g''(t)=t/sqrt(4-t)-t/(4(4-t)^(3/2))$

$g''(8/3)=-3sqrt(3)/4<0$
so we get $PMAX(8/3;16*sqrt(3)/9)$

How do you find the critical numbers for g(t) = t sqrt(4-t) g(t)=t√4−tg(t) = t sqrt(4-t) to determine the maximum and minimum?