


There is one critical point at (2, 4), but this point happens to fall on one of the boundaries of the region. We'll get to that point in a moment.
Along the boundary
, we have

which attains a maximum value of

Along
, we have

which attains a maximum of

Along
, we have

which attains a maximum of

So over the given region, the absolute maximum of
is 1578 at (2, 44).
We can start by finding the gradient of LM

Two perpendicular lines will meet the requirement

×

=-1
Two parallel lines have equal gradients
NM is perpendicular to LM, hence the gradient of NM is -1
KN is a line that is parallel to NM, hence the gradient is 1
KL is perpendicular to LM, hence the gradient of KL is -1
What kind of question is that
Answer:
≈ 11.78 in
Step-by-step explanation:
The arc length is calculated as
arc = circumference of circle × fraction of circle
= 2πr ×
( cancel 2π on numerator/ denominator )
= 15 × 
= 
≈ 11.78 in ( to 2 dec. places )