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Consider all points in the plane that solve the equation \(x^2+2y^2=16\). Find the maximum value of the product \(xy\) on this graph.

 

(In case the LaTeX doesn't render)--Consider all points in the plane that solve the equation x^2 + 2y^2 = 16. Find the maximum value of the product xy on this graph.

 Jan 26, 2020
 #1
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The maximum value is 16/3, which you get when x = y = 4/sqrt(3).

 Jan 26, 2020
 #2
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"The maximum value is 16/3, which you get when x = y = 4/sqrt(3)." 

 

This is wrong. Show steps.

Guest Jan 27, 2020
 #3
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+1

Explain why this is wrong.

Guest Jan 27, 2020
 #4
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My teacher said it was wrong.

 

Explain why it is right. Show steps.

Guest Jan 27, 2020
 #6
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Your teacher is wrong.  The answer is 16/3.

Guest Jan 27, 2020
 #9
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I do not believe that the answer is 16/3

Melody  Jan 27, 2020
 #10
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x and y will not be equal

Melody  Jan 27, 2020
 #5
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Consider all points in the plane that solve the equation x^2 + 2y^2 = 16. Find the maximum value of the product xy on this graph.

 

Doing it graphically I get approx  5.66

 

 Jan 27, 2020
 #7
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I really need the answer!!!

 Jan 27, 2020
 #8
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I gave you an answer, I do not see your thanks.

 

It is easy enough to get an exact answer using calculus.

 

\(x=\sqrt{16-2y^2}\\ let \;\;M=xy\\ M=y*\sqrt{16-2y^2}\\ Find \quad dM/dy\\ \text{put it equal to 0 to find the stationary points.}\\ \text{Solve for y}\\ \text{Substitute to find M}\\ \text{if you are feeling keen you should prove that this is a maximum.}\\ \text {This will give you the exact value that i have already approximated via the graph.}\\ \)

Melody  Jan 27, 2020
 #11
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Easiest is to use the parametric equations of the ellipse.

 

For the ellipse

 \(\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, \\\text{ the parametric equations are }\\x = a\cos \theta,\qquad y = b\sin \theta.\)

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 Jan 27, 2020
 #12
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Thanks Tiggsy,

I thought there was probably a better way. Or at least a way that did not involve graphing or calculus.  :)

Melody  Jan 27, 2020

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