What is the area of the trapezoid
2 answers:
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Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
Answer:
(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.
Step-by-step explanation:
Given that all the boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over.
This implies that boxes are in multiples of 12
i.e. 12, or 24, or 36 or....
Let us say this as 12m for m an integer
Now additional boxes arrived =60
Total number of boxes now =
This when arranged in stacks of 14,nothing left out
Or this is multiple of 14
for some positive integer n.
By trial and error we find that if 60 is to be divided by 14, 24 should be added and 24 is a multiple of 12.
So we say originally 24 boxes were there now 84 boxes there.
Hence both options are right.
(1) There were fewer than 110 boxes in the warehouse before the 60 additional boxes arrived. (2) There were fewer than 120 boxes in the warehouse after the 60 additional boxes arrived.