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Calculus/Max Area
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Question
A little help
What is the Max Area of a redtangle inscribe in the ellipes x^2+4y^2=4 with the sides parallel to the coord axes.
I set up y=sqrt((4-x^2)/(4))
max = 7.6
an I correct
Hello Georgia,
Yes, that is a good start!
Actually, y=(+/-)(1-x^2/4)^(1/2), i.e. plus or minus...
So the dimensions of the rectangle will be
2x by 2sqrt(1-x^2/4)^(1/2)...for an area of
A(x)=4xsqrt(1-x^2/4)...
Now solve dA/dx=0...which yields: x=sqrt(2)
For a max. area of 4 square units...OK?
TTYL, Abe
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Answers by Expert:
Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience
Over 15 years teaching at the college level.
Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.
Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook
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