A defined on an interval, then the following holds.

If $A^{\text{'}}(w)>0$ for all

$A^{\text{'}}(w)=w=w$ and $A^{\text{'}}(w)>0$ for all $w>c,$ then $A(c)is$ the $absolute$ minimum value $ofA.$

$To$ apply this test $we$ must first find $A^{\text{'}}(w).$ Doing $so$ plyes the following result.

$A(w)=-w^{2}36w$

$A^{\text{'}}(w)=$

Setting this derivative equal $to0$ and solving for $w$ gives the following result.

$w=$

$$

Applying the test, $we$ see that there $isanabsoluteat$ this value $ofw.w$ and $A^{\text{'}}(w)<mn>0</mn>$ for all $w>c,$ then $A(c)is$ the $absolute$ maximum value $ofA.$

$If{A}^{\text{'}}(w)<mn>0</mn>$ for all $w$ and $A^{\text{'}}(w)>0$ for all $w>c,$ then $A(c)is$ the $absolute$ minimum value $ofA.$

$To$ apply this test $we$ must first find $A^{\text{'}}(w).$ Doing $so$ plyes the following result.

$A(w)=-w^{2}36w$

$A^{\text{'}}(w)=$

Setting this derivative equal $to0$ and solving for $w$ gives the following result.

$w=$

$$

Applying the test, $we$ see that there $isanabsoluteat$ this value $ofw.$