Expert: Sherman D. - 8/21/2006

Question
question 3, 4, and 5, i worked out question 1 and 2.

The upper half of the ellipse
[(x^2)/25] + [(y^2)/16]= 1
with y>=0, contains a rectangle with vertices A(x, y), B(−x, y),C(−x, 0), D(x, 0), where (x, y) is a point on the ellipse in the first quadrant.

(i) Draw a sketch of the upper half of the ellipse and the inscribed rectangle.
DONE:          
         |
       (-x,y) ______|_______(x,y)
         |     |     |
         |     |     |
     --------------------------
         (-x,0)        (x,0)
with half cirlce/ellipse

(ii) Show that the area of the rectangle is given by A = (8/5)*(x)*sqrt(25 − x^2).
DONE:
x^2= [(400)-(25)(y^2)]/16
y^2= [(400)-(16)(x^2)]/25
both from
[(x^2)/25] + [(y^2)/16]= 1

A=2xy
= 2x*sqrt([(400)-(16)(x^2)]/25)
= (8/5)*(x)*sqrt(25 − x^2).


(iii) Show that for any fixed value of A < 20, there are two possible rectangles that will fit inside the upper half of the ellipse.

(iv) Find the dimensions of the two rectangles which have area A =91/5.
(please check my working)
91/5=(8/5)*(x)*sqrt(25 − x^2)
91=8x*sqrt(25 − x^2)
(squared both sides)
91^2=(8x)^2*(25
91^2= 1600x^2-64x^4
64x^4-1600x^2+91^2=0
let z=x^2
64z^2-1600z+91^2=0
using [-b^2(+/-)sqrt((b^2)-4ac)]/2a
[-(1600^2)(+/-)srqt((1600^2)-4(64)(91^2))]/2(64)
z=[12 1/2 (+/-) 3/8(sqrt(191))]
x=sqrt(z)
=4.205...
or
=2.705...
is this right?

(v) Find the dimensions of the rectangle of largest area that will fit into the upper half of the ellipse and
give its area.


Answer
As far as #1 goes, go to http://www.calculator.com/calcs/GCalc.html
type in 4sqrt(1 - ((x^2)/25)), but leave out the y = part, and don't worry about also typing it -4sqrt(1 - ((x^2)/25))
because you only need one side of the ellipse and not both.

Now plot points at (x,y), (-x,y), (x,0), and (-x,0) and you will have your rectangle

To get the second one, just plot points that instead of going from right to left the long way, it goes from up to down the long way.

-------------------------------

y = 4sqrt(1 - ((x^2)/25))
y = 4sqrt((25/25) - ((x^2)/25))
y = 4sqrt((25 - x^2)/25)
y = (4/5)sqrt(25 - x^2)

Thats why i got for "y"

Here is why Area equals (8/5) * x * sqrt(25 - x^2)

Area = xy

In most cases.

But since "x" goes from the point on the half-ellipse to the x-axis, thats just a solid width.

Now as for the "y", since y = (4/5)sqrt(25 - x^2) only measures from the y-axis to the point on the half-ellipse, in order to get the full length, you will have to double the "y" value, so you now have

Area = 2xy

So this gives you

Area = 2 * ((4/5)sqrt(25 - x^2)) * x
Area = (8/5)sqrt(25 - x^2) * x

So its not really the point of prooving it, its just common since.

------------------------

Don't know if this is prooving it or not

But the 2 rectangles have an area of

A1 = (8/5)sqrt(25 - x^2) * x
A2 = (8/5)sqrt(25 - y^2) * y

You have a vertical and horizontal rectangle

If you double the length of y or x and multiply it by x or y respectively, you will get a maximum area of 20.

for a graph, go to www.calculator.com/calcs/GCalc.html
type in (4/5)sqrt(25 - x^2), but leave out the y = part or the graph will not show up.