Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
Answer:
Kiaris I did it check ur Gamil
Step-by-step explanation:
Anddd Similar traingles for one more time.
Here,
RQ/RP = RS/RT = QS/PT = 1/2
Using the values of QS and PT will give,
Therefore y= 46 units.
Answer: -4.25 ≤ x ≤ 6.25
Step-by-step explanation:
The midpoint of a segment is located at the same distance of each of the endpoints of the segment.
The midpoint is at x = 1.
The length of the segment is 10.5, if we divide it by two we have:
10.5/2 = 5.25
Now, if we want the endpoints to be at the same distance of the midpoint, then the endpoints will be:
Xmax = midpoint + 5.25
Xmin = midpoint - 5.25
Then the extremes are:
Xmax = 1 + 5.25 = 6.25
Xmin = 1 - 5.25 = -4.25
Then this segment can be written as:
Xmin ≤ x ≤ Xmax
-4.25 ≤ x ≤ 6.25
Answer:
yes
Step-by-step explanation:
i dont know
