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ellipse optimization?
Find the dimensions of the largest rectangle that can be inscribed in the ellipse= (x^2)/4 +(y^2)/9 =1
1 Answer
- SqdancefanLv 76 years ago
The equation can be solved for y to get
.. y = √(9*(1 - x^2/4)) = (3/2)√(4-x^2)
The product x*y will be the area of 1/4 of the rectangle. We can find the maximum by setting the derivative of this expression to zero
.. d(xy)/dx = 0 = (3/2)(√(4-x^2) + x(1/2)/√(4-x^2)*(-2x))
.. = 3/(2√(4-x^2))*(4 - x^2 - x^2)
.. = 3(2 - x^2)/√(4-x^2)
This is zero when
.. 2 - x^2 = 0
or x = ±√2
For this value of x, we have
.. y = (3/2)√(4-x^2) = (3/2)√2
The largest rectangle that will fit in the ellipse is 2√2 wide by 3√2 tall.
