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The given ellipse is x2/a2 + y2/b2=1, that means that it is 2a wide as well as 2b tall
In the respective diagram which has been provided, it can be noticed that x is considered to be half width of rectangle and y is considered to be the half the height. Therefore, in such case, the area of the rectangle is:
A= (2x) (2y) = 4xy
In order to maximise the same, there should be solving of ellipse equation for y and the substitute for y is considered to be in area equation. Solving of ellipse of y, there can be usage of the positive square root that can be represented the upper half of ellipse:
y2/b2 = 1- x2/a2
y/b = √1-x2/a2
y= b (√1/a2 (a2-x2)
= b/a √a2-x2
Moreover, the substitute for y in rectangle area equation and finding the derivative
A= 4xy = 4x (b/a √a2 – x2) = 4b/ax √a2 – x2)
A’ = 4b/a ((1) √a2- x2 + x * ½ (a2 – x2) -1/2 (-2x))
= 4b/a (√a2-x2 – x2/√a2 – x2)
= 4b/a (a2 – x2/√a2- x2 – x2/√a2- x2)
= 4b/a (a2- 2x2/ √a2-x2)
Now, finding wherein A’= 0 and in such scenario, the numerator is 0
0 = a2 – 2x2
X √2 = a
Therefore, from the same, it can be confirmed that it is local maximum of A, however, as the minimum area would be zero, it is likely to be seen that it is the maximum as well.
Now, finding corresponding y value with ellipse equation from the overall above are as follows:
Y= b/a √a2- x2
= b/a √a2 – (a/√2)2
=b/a √2a2/2 – a2/2
= b/a √a2/2
= b/a* a/√2
Now, plugging both x as well as y in the rectangle area related equation is:
A= 4xy = 4* a/√2 * b/√2 = 4ab/2 = 2ab
Therefore, the result is 2ab.
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