math anyone? calculus?

Question

Find the dimensions of the rectangle of largest area that can be inscribed in the ellipse (x^2)/(a^2) + (y^2)/(b^2) = 1. (the answer may only depend on a and b and assume a is less than or equal to b).

Answer 1

the width of the rectangle is 2x
height = 2y where y = b/a(sqrt(a^2-x^2)
area = 4xy = (4bx/a)sqrt(a^2-x^2)
differntiating and setting the derivative = 0 , we get
sqrt(a^2-x^2) - x^2/sqrt(a^2-x^2)=0
solving for x, we get x = a/sqrt2
substituing and solving for y we get y = b/sqrt2
thus dimensions of rectangle required are
width=asqrt2 and height = bsqrt2
i leave it to you to verify it is a max for those values.

Answer 2

Let the top right vertice of the rectangle be A(h,k)
=> the one opposite to it horizontally is B(-h,k), the one below B on same side is C(-h,-k) and last is D(h,-k).

Area(S) of rectangle will be=4(hk)
Because of symmetry in each quadrant

=>S=4(h)sqrt[{1-(h/a)^2}b^2]

Since Area is alway a positive quantity the differential will not be affected on squaring S

=>T=S^2=16h^2(1 - (h/a)^2)b^2

=>T=16(hb)^2 - 16(bh^2/a)^2

=>dT/dh=32hb^2 - 64h^3 * (b/a)^2

For maximum dT/dh=0

=>32hb^2=64h^3 * (b/a)^2

=>1=2h^2/a^2

=>h=a/sqrt(2)

=>k=sqrt(1-a^2/(2a^2))b^2

=>k=b/sqrt(2)


=>Using formula, dimensions are:
sqrt((a/sqrt(2) - a/sqrt(2))^2 + (b/sqrt(2) + b/sqrt(2))^2)

=>sqrt(bsqrt(2))^2

=>bsqrt(2)

and sqrt((a/sqrt(2) + a/sqrt(2))^2 + (b/sqrt(2) - b/sqrt(2))^2)

=>asqrt(2)


=>b*sqrt(2), a*sqrt(2)

Answer 3

If the rectangle extends from -x to +x in the horizontal direction, it will extend from -b√(1-x^2/a^2) to +b√(1-x^2/a^2) in the vertical direction. So the area is
A(x) = 2x (2b√(1-x^2/a^2))
= (4b/a) x √(a^2-x^2)
A'(x) = (4b/a) [√(a^2-x^2) + x (1/2)(a^2-x^2)^(-1/2) (-2x)]
= (4b/a) (a^2-x^2)^(-1/2) [a^2 - x^2 - x^2]
= 0 when x = a/√2
So the width is 2a/√2 = a√2 and the height is 2b√(1-x^2/a^2) = b√2, for a total area of ab/2.

Answer 4

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