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# Ellipse problem-help

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

Melody  Jan 27, 2020
#10
+108624
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x and y will not be equal

Melody  Jan 27, 2020
#5
+108624
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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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Jan 27, 2020
#8
+108624
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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