Answer:
The possible dimensions of the deck can be anything that adds up to be 130 m.
Step-by-step explanation:
<em>The perimeter could be 1m x 64m.</em>
Two sides would be 1m each (so 2m total) and the other two sides would be 64m each (128m total). 2 + 128 = 130m.
<em>The perimeter could be 15m x 50m.</em>
Two sides would be 15m each (so 30m total) and the other two sides would be 50m each (100m total). 30 + 100 = 130m.
2+3+7+8=20
2 x 3+7+ 8= 21
2 x 7 + 3+ 8= 25
8 x 2 + 7 + 3= 26
3 x 7 +8 - 2= 27
8 x 3 + 7 - 2= 29
8 x 7 ÷ 2 + 3= 31
7 x 2 x 3 -8 = 34
There is a couple, but i am not doing all of them
I hope this helps anyway :).
Let x represent the side length of the square end, and let d represent the dimension that is the sum of length and girth. Then the volume V is given by
V = x²(d -4x)
Volume will be maximized when the derivative of V is zero.
dV/dx = 0 = -12x² +2dx
0 = -2x(6x -d)
This has solutions
x = 0, x = d/6
a) The largest possible volume is
(d/6)²(d -4d/6) = 2(d/6)³
= 2(108 in/6)³ = 11,664 in³
b) The dimensions of the package with largest volume are
d/6 = 18 inches square by
d -4d/6 = d/3 = 36 inches long
<span>B) Observe several cases of people ordering food of varying spice-levels and number of soft drinks ordered. thats what i think though.</span>
The nearest whole number would be 96