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Find the greatest and smallest values that the function
f(x, y) = xy
takes on the ellipse
x2/8 + y2/2= 1?
f(x, y) = xy
takes on the ellipse
x2/8 + y2/2= 1?
Most Upvoted Answer
Find the greatest and smallest values that the functionf(x, y) = xytak...
Understanding the Problem
To find the extrema of the function f(x, y) = xy on the ellipse defined by the equation x²/8 + y²/2 = 1, we can use the method of Lagrange multipliers or parameterization.
Parameterization of the Ellipse
The ellipse can be parameterized as follows:
- x = 2√2 * cos(t)
- y = √2 * sin(t)
Where t varies from 0 to 2π.
Expression of f(x, y)
Substituting the parameterization into f(x, y):
- f(t) = (2√2 * cos(t)) * (√2 * sin(t)) = 4sin(t)cos(t)
- This simplifies to f(t) = 2sin(2t)
Finding Extrema
To find the maximum and minimum values:
- The function 2sin(2t) oscillates between -2 and 2.
Maximum and Minimum Values
- The greatest value of f(x, y) on the ellipse is 2.
- The smallest value of f(x, y) on the ellipse is -2.
Conclusion
Therefore, the extrema of the function f(x, y) = xy on the ellipse x²/8 + y²/2 = 1 are:
- Greatest value: 2
- Smallest value: -2
To find the extrema of the function f(x, y) = xy on the ellipse defined by the equation x²/8 + y²/2 = 1, we can use the method of Lagrange multipliers or parameterization.
Parameterization of the Ellipse
The ellipse can be parameterized as follows:
- x = 2√2 * cos(t)
- y = √2 * sin(t)
Where t varies from 0 to 2π.
Expression of f(x, y)
Substituting the parameterization into f(x, y):
- f(t) = (2√2 * cos(t)) * (√2 * sin(t)) = 4sin(t)cos(t)
- This simplifies to f(t) = 2sin(2t)
Finding Extrema
To find the maximum and minimum values:
- The function 2sin(2t) oscillates between -2 and 2.
Maximum and Minimum Values
- The greatest value of f(x, y) on the ellipse is 2.
- The smallest value of f(x, y) on the ellipse is -2.
Conclusion
Therefore, the extrema of the function f(x, y) = xy on the ellipse x²/8 + y²/2 = 1 are:
- Greatest value: 2
- Smallest value: -2
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Find the greatest and smallest values that the functionf(x, y) = xytakes on the ellipsex2/8 + y2/2= 1?
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Find the greatest and smallest values that the functionf(x, y) = xytakes on the ellipsex2/8 + y2/2= 1? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Find the greatest and smallest values that the functionf(x, y) = xytakes on the ellipsex2/8 + y2/2= 1? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the greatest and smallest values that the functionf(x, y) = xytakes on the ellipsex2/8 + y2/2= 1?.
Find the greatest and smallest values that the functionf(x, y) = xytakes on the ellipsex2/8 + y2/2= 1? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Find the greatest and smallest values that the functionf(x, y) = xytakes on the ellipsex2/8 + y2/2= 1? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the greatest and smallest values that the functionf(x, y) = xytakes on the ellipsex2/8 + y2/2= 1?.
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