Question
Find the rectangle with the maximum area inside the ellipse x^2/16 + y^2/4=1.
Thanks.
Thanks.
Answers
If the base of the rectangle has length 2x, and height y, the area is
a = 2xy = 2x√(1-x^2/16)(4)
= 2x√(16-x^2)
da/dx = 4(8-x^2)/√(16-x^2)
so, max area occurs at x=√8, and the rectangle is 2√8 by √2
a = 2xy = 2x√(1-x^2/16)(4)
= 2x√(16-x^2)
da/dx = 4(8-x^2)/√(16-x^2)
so, max area occurs at x=√8, and the rectangle is 2√8 by √2
Related Questions
The area of an ellipse in the form (x*x/a*a)+(y*y)/(b*b)=1 is A=3.14*ab. For this ellipse, a+b=20....
write a quadratic function needed to find the dimension of the rectangle with perimeter 100 having t...
find the dimensions of the rectangle of maximum area that can be inscribed inside the ellipse (x/4)^...
Find the dimensions of a rectangle of maximum area with sides parallel to the coordinate axes that c...