a. Note that
is continuous for all
. If
attains a maximum at
, then
. Compute the derivative of
.

Evaluate this at
and solve for
.




To ensure that a maximum is reached for this value of
, we need to check the sign of the second derivative at this critical point.

The second derivative at
is negative, which indicate the function is concave downward, which in turn means that
is indeed a (local) maximum.
b. When
, we have derivatives

Inflection points can occur where the second derivative vanishes.




Then we have three possible inflection points when
,
, or
.
To decide which are actually inflection points, check the sign of
in each of the intervals
,
,
, and
. It's enough to check the sign of any test value of
from each interval.




The sign of
changes to either side of
and
, but not
. This means only
and
are inflection points.
First you start off by setting up a Subtraction problem there anything common denominators of 7/8 and two over three verse 7/8 you should get 21/24 invert to over three you should get 16/24 then you would subtract those two anyway to get 5/24 I hope that was helpful get it
The given polynomial is
→ 6 ac - 15 a d- 8 b c + 2 b d
→ 3× 2×a×c - 3×5×a×d - 2×2×2× b×c + 2×b×d
→ 3 a ( 2 c -5 d) - 2 b(4 c -d)
There would be nine lilies and 4 tulips.
hope it helps :)