Question

A geometry student wants to draw a rectangle inscribed in the ellipse x^2 +4y^2 =9. What is the area of the largest rectangle that the student can draw?

Answer 1

Let the lengths of the sides of rectangle be a*2 and b*2 parallel to x axis and y axis respectively;
a^2 + 4b^2 = 9
b = (9- a^2)/4
Let the area of the rectangle be a function of a:
A(a) = a*2*(9- a^2)/4*2
= 9a - a^3
you will want to find the local extrema of A(a),
that is,
dA/da = 0
9 - 3a^2 = 0
a = sqrt(3) or -sqrt(3)

Answer 2

a second order derivative test shows that A''(sqrt(3)) < 0, A''(-sqrt(3)) > 0, therefore A(sqrt(3)) is a local maxima.
The area of the rectangle is therefore 9*sqrt(3) - (sqrt(3))^3

Answer 3

Why are the side lengths equal to \[a ^{2} and. b ^{2}\]?

Answer 4

not a^2 , a*2

Answer 5

OH. That makes a lot more sense XD Thanks!

Answer 6

If a^2 + 4b^2 = 9
Shouldn't b = sqrt( (9- a^2)/4 ) ?

Answer 7

I am getting:

\[a = \frac{ 3\sqrt{2} }{ 2 }\]
\[b = \frac{ 3\sqrt{2} }{ 4 }\]

Answer 8

b = sqrt((9- a^2)/4)
Let the area of the rectangle be a function of a:
A(a) = a*2*((9- a^2)/4)^(1/2)*2
= a*2*((9- a^2)*(1/2))
you will want to find the local extrema of A(a),
that is,
dA/da = 0
2*(9-a^2)^(1/2)-2*a^2*(9-a^2)^(-1/2)=0
a = 3/sqrt(2)
b = 9/8

Answer 9

I am getting Maximum Area = 9

Area = 4ab with a and b given in my earlier reply.

Answer 10

DX the last line of my solution was supposed to be b^2
I should really be more careful with these trivial things.

Answer 11

yeah that is b^2.
Your a and mine are the same if you multiply your numerator and denominator by sqrt(2)
And if you take the square root of 9/8 your b will be the same as mine.
Some textbooks and teachers don't like radical signs in the denominator and so I get rid of them in the denominator.