SOLUTION: Find the area of the largest possible rectangle that can inscribe in an ellipse. 9x^2 + 4y^2 = 36.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the area of the largest possible rectangle that can inscribe in an ellipse. 9x^2 + 4y^2 = 36.       Log On


   

Question 1082936: Find the area of the largest possible rectangle that can inscribe in an ellipse. 9x^2 + 4y^2 = 36.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
I'll solve the 1/4 problem and then multiply by 4.
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A=xy
Define a new function which is the square of A,
Z=A%5E2=x%5E2y%5E2
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Make Z a function of one variable using the ellipse,
4y%5E2=36-4x%5E2
y%5E2=%2836-9x%5E2%29%2F4
Substituting,
Z=%28x%5E2%2836-9x%5E2%29%29%2F4
Take the derivative of Z with respect to x and set it equal to zero.
dZ%2Fdx=-9x%28x%5E2-2%29=0
x=0 is the trivial solution so,
x%5E2=2
x=sqrt%282%29
So then solving for y%5E2,
9%282%29%2B4y%5E2=36
4y%5E2=18
y%5E2=9%2F2
y=3%2Fsqrt%282%29%7D%7D%0D%0A%7B%7B%7By=%283%2F2%29sqrt%282%29
So then the maximum area would be,
A%5Bmax%5D=4%28sqrt%282%29%29%283%2Fsqrt%282%29%29=12
The width of the rectangle (x direction) would be,
W=2sqrt%282%29
and the length of the rectangle (y direction) would be,
L=2%283%2F2%29sqrt%282%29=3sqrt%282%29