Angel R.

asked • 11/05/14

What is the maximum area of a rectangle?

that is contained in a semi ellipse whose major axis is 12 and minor axis is 8. how would you solve this? I got the area as 24 is it right?

Christopher R.

Let's write the general equation of an ellipse.
 
 x^2/a^2 + y^2/b^2 =1
-x^2/a^2                   -x^2/a^2
 
y^2/b^2=1-x^2/a^2 = (a^2 - x^2)/a^2
 
This implies y^2 = b^2/a^2 *(a^2 - x^2)
Hence,
 
y=b/a*√(a^2-x^2) Note: this is the equation of a semi ellipse. The other half is y=-b/a*√(a^2-x^2) below the x-axis.
 
The problem states the rectangle is within the semi ellipse where its height is y and length is 2x. This implies its area can be written as a function of x, which is:
 
A(x)=2xy = 2x*b/a*√(a^2-x^2)
 
Take the derivative of A(x) with respect to x and set it to equal to zero.
 
A'(x)=0=2b/a*(√(a^2-x^2) + -2x*1/2*x/√(a^2-x^2)) =√(a^2-x^2) - x^2/√(a^2-x^2) Note: 2b/a drops out due to being the factor outside of the equation and it being equal to zero.
 
Multiply the equation by √(a^2-x^2) in which you get:
 
0=a^2 - x^2 - x^2 = a^2-2x^2 This implies 2x^2=a^2
 
Hence, xmax=a/√2
 
A(xmax) = 2a/√2*b/a*√(a^2-a^2/2) = 2b/√2*a*√2 = 2ba=2ab
 
Therefore, Tom F. was correct in determining the maximum area of the rectangle being A=2ab
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11/05/14

Christopher R.

Sorry, I made a simple error in my calculation at the end in proving the formula in determining the maximum area of the rectangle.
 
A(xmax)=2b/√2*a/√2 = 2ba/2 = ba =ab Therefore, the maximum area of the rectangle within the semi ellipse is
A=ab. Again, you were correct in your original answer, where a=6 and b=4 which makes the area being 24 sq units. 
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11/05/14

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Christopher R. answered • 11/05/14

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First of all, write out the general equation of the given ellipse in which is:
 
x2/122+y2/82=1
 
With some algebraic manipulation:
 
y=2/3*√(144-x2)
 
Let A(x) equal the area of the rectangle contained in the semi ellipse be a function of x.
 
A(x)=2xy=2x*2/3*√(144-x2)
 
A(x)=4/3*x*√(144-x2)
 
Take the derivative of A(x) with respect to x
 
A'(x)=4/3*√(144-x2)+4/3*x*1/2*(-2x)*(144-x2)-1/2
 
A'(x)=0=4/3*(√(144-x2)-x2/√(144-x2))
 
Multiply both sides of the equation by √(144-x2) in which you get:
 
144-x2-x2=0 This implies 2x2=144 Hence, x = √72 ≈8.485 
 
A(√72)=4/3*√72*√(144-72)=4/3*√72*√72=4/3*72=96 units2 is the maximum area of the rectangle
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Angel R.

wouldn't the general equation be (x^2/6^2)+(y^2/4^2)=1
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11/05/14

Christopher R.

Yeah, your right, If the major axis is 12 units a=6 and minor axis being 8 would make b=4. Sorry, I made the mistake of not taking half of the given numbers to get a and b. I did some calculations and found the maximum area of the rectangle being Amax=ab
 
Thus, your original answer is correct being 24.
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11/05/14

Angel R.

OK thank you sir =]
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11/05/14

Tom F.

tutor
I think you are close here but not correct.
 
In your Area formula the correct expression should be A =2x*2y
 
If you use that and use a = 6 and b =4 and then do the same work you will see that the correct answer is in fact 48
 
Give it a try and I think you will see what I mean
 
Good Luck
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11/05/14

Tom F.

tutor
the correct function for A is
 
A = 16x SQRT(1- x2/36)
 
Take the derivative of this and solve and you will be all set
 
 
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11/05/14

Tom F. answered • 11/05/14

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The maximum area of the rectangle is equal to 2ab
 
In this case a = 6  and b = 4
 
The work shown in the other answer should lead to this result if you use 6 and 4 in the general form of the equation
 
maximum area
 
=2*6*4
 
= 48 sq units
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Christopher R.

Where did you get a=6 and b=4?
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11/05/14

Angel R.

I did the same process he did over and over again but i keep getting 24. Fro my critical point i got 3sqrt(2). I plugged it into my equation of the semi-ellipse and got a y value of 4sqrt(1/2) , then plugged those values into A=2xy and get 24. What am i doing wrong?
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11/05/14

Angel R.

isnt it t since the major axis is 12, you go 6 units to the right from the origin then 6 units to the left? same thing with minor axis , 4 up and 4 down.
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11/05/14

Tom F.

tutor
I did the same process he did over and over again but i keep getting 24. Fro my critical point i got 3sqrt(2). I plugged it into my equation of the semi-ellipse and got a y value of 4sqrt(1/2) , then plugged those values into A=2xy and get 24. What am i doing wrong?
 
 
YOU DID EVERYTHING CORRECTLY! 
 
The Area formula should be A = 2x*2y   using this you would get the correct answer of 48
 
Good luck
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11/05/14

Angel R.

Isnt it a=2xy because its half of an ellipse? If it were the full ellipse it would be a= 2x2y?
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11/05/14

Tom F.

tutor
No, it is 
 
Area = length * width
 
Length = 2x  and width = 2y
 
then 
 
A = 2x * 2y  it has nothing to do with the ellipse
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11/05/14

Tom F.

tutor
So sorry,  you are right.  I just realized that we were talking about half of the ellipse.
 
All of my answers were for the full ellipse.
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11/05/14

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