Triangle XYZ has points X = (–4,4), Y = (0,1), and Z = (4,4). Triangle X'Y'Z' has points X ’= (–5,–4), Y’ = (–1,–1), Z’ = (3,–4)
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Answer:
The maximum variance is 250.
Step-by-step explanation:
Consider the provided function.
Differentiate the above function as shown:
The double derivative of the provided function is:
To find maximum variance set first derivative equal to 0.
The double derivative of the function at is less than 0.
Therefore, is a point of maximum.
Thus the maximum variance is:
Hence, the maximum variance is 250.