Given the functions f(×)=-×2+9 and g(×)=6-2×
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Answer:
A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds. 4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds
Step-by-step explanation:
The given table is presented as follows;
The number of laps in the range 82 to 84 seconds = 1
The number of laps in the range 84 to 86 seconds = 4
The number of laps in the range 86 to 88 seconds = 2
The number of laps in the range 88 to 90 seconds = 4
The number of laps in the range 90 to 92 seconds = 6
The number of laps in the range 92 to 94 seconds = 5
The number of laps in the range 94 to 96 seconds = 2
The number of laps in the range 96 to 98 seconds = 0
Therefore, the histogram that represents Blanca's lap times for the three days of practice is described as follows;
A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds. 4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds
I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to 
.
Set
and 
, so that the Lagrangian is
Take the partial derivatives and set them equal to zero.
One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives
So by adding the first three equations together, you end up with
and the fourth equation allows you to write
Now, substituting this into the first three equations in the most recent system yields
So we found a grand total of 8 possible critical points. Evaluating at each of these points, you find that 
attains a maximum value of 
whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of 
whenever exactly one or all of the critical points' coordinates are negative.
Set
Take the partial derivatives and set them equal to zero.
One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives
So by adding the first three equations together, you end up with
and the fourth equation allows you to write
Now, substituting this into the first three equations in the most recent system yields
So we found a grand total of 8 possible critical points. Evaluating