The function you seek to minimize is
()=3‾√4(3)2+(13−4)2
f
(
x
)
=
3
4
(
x
3
)
2
+
(
13
−
x
4
)
2
Then
′()=3‾√18−13−8=(3‾√18+18)−138
f
′
(
x
)
=
3
x
18
−
13
−
x
8
=
(
3
18
+
1
8
)
x
−
13
8
Note that ″()>0
f
″
(
x
)
>
0
so that the critical point at ′()=0
f
′
(
x
)
=
0
will be a minimum. The critical point is at
=1179+43‾√≈7.345m
x
=
117
9
+
4
3
≈
7.345
m
So that the amount used for the square will be 13−
13
−
x
, or
13−=524+33‾√≈5.655m
Answer:
the answer is -2.5
Step-by-step explanation:
Answer:
height
Step-by-step explanation:
you can just look these questions up
See the attachment for detailed answer
<span>So then R is 1/6 of the way from E to F.
The x distance from E to F is 11- 4=7.
So then the x distance from E is,
Similarly for y,
R:(31/6,22/3)</span>