Honeycomb The surface area of a cell in a honeycomb is S = 6 h s + 3 s 2 2 ( 3 − cos θ sin θ ) where h and s are positive constants and θ is the angle at which the upper faces meet the altitude of the cell (see figure). Find the angle θ ( π / 6 ≤ θ ≤ π / 2 ) that minimizes the surface area S .
Honeycomb The surface area of a cell in a honeycomb is S = 6 h s + 3 s 2 2 ( 3 − cos θ sin θ ) where h and s are positive constants and θ is the angle at which the upper faces meet the altitude of the cell (see figure). Find the angle θ ( π / 6 ≤ θ ≤ π / 2 ) that minimizes the surface area S .
Honeycomb The surface area of a cell in a honeycomb is
S
=
6
h
s
+
3
s
2
2
(
3
−
cos
θ
sin
θ
)
where h and s are positive constants and
θ
is the angle at which the upper faces meet the altitude of the cell (see figure). Find the angle
θ
(
π
/
6
≤
θ
≤
π
/
2
)
that minimizes the surface area S.
For the given graph, determine the following.
-3
12
УА
4
3
-
-1
°
1 2
3
x
-1.
-2-
a. Determine for which values of a the lim f (x) exists but f is not continuous at x = a.
a
b. Determine for which values of a the function is continuous but not differentiable at x = a.
a
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.