Please help me with this problem.
1 answer:
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The dimensions of the square base after cutting the corners will be (40-2x). The depth of the box will be x, so its volume is given by
V = x(40-2x)²
You can differentiate this to get
V' = 12x² -320x -1600
Setting this to zero and factoring gives
(3x-20)(x-20) = 0
The appropriate choice of solutions is
x = 20/3 = 6 2/3
The dimensions of the box of maximum volume are
26 2/3 in square by 6 2/3 in deep
The maximum volume is
(80/3 in)²(20/3 in) = 4740 20/27 in³
V = x(40-2x)²
You can differentiate this to get
V' = 12x² -320x -1600
Setting this to zero and factoring gives
(3x-20)(x-20) = 0
The appropriate choice of solutions is
x = 20/3 = 6 2/3
The dimensions of the box of maximum volume are
26 2/3 in square by 6 2/3 in deep
The maximum volume is
(80/3 in)²(20/3 in) = 4740 20/27 in³
The answer is:
The rate of change is not constant and increases then decreases over time. The height of the ball above ground gets larger until 1.25 seconds and then gets smaller after that time.
Here's how:
The rate of change of the function is defined and calculated as (refer to the statement beloew):
r = [change in height] / {change in time]
For the Table:
refer to the attached picture.
The table shows the calculations for the rate of change (r) for each interval given.
And for the Conclusion,
Refer to the table and notice that in the third ans fifth columns show that:
The rate of change is not constant and increases then decreases over time. The height of the ball above ground gets larger until 1.25 seconds and then gets smaller after that time.
The rate of change is not constant and increases then decreases over time. The height of the ball above ground gets larger until 1.25 seconds and then gets smaller after that time.
Here's how:
The rate of change of the function is defined and calculated as (refer to the statement beloew):
r = [change in height] / {change in time]
For the Table:
refer to the attached picture.
The table shows the calculations for the rate of change (r) for each interval given.
And for the Conclusion,
Refer to the table and notice that in the third ans fifth columns show that:
The rate of change is not constant and increases then decreases over time. The height of the ball above ground gets larger until 1.25 seconds and then gets smaller after that time.