Answer:
Step-by-step explanation:
m ∠XOY = m ∠WOV
so, m XY = m WV --- (i)
{ If angle subtended by two arcs at the center are equal, then length of arc are equal}
m YZ = m ZW ------- (ii) {given}
Add (i) and (ii)
XY + YZ = WV + ZW
XZ = ZV
Hence proved.
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:
Let w be the width. Then the length is 2w-4. So:
w(2w-4)=96
2w²-4w-96=0
w²-2w-48=0
(w-8)(w+6)=0
w=8 or -6
Throwing out the negative value for w, we get a width of 8 yds, and a length of 12 yds. ☺☺☺☺
