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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?
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
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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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