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

Guest Jan 23, 2018
 #1
avatar+26829 
+2

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

 

.

Alan  Jan 23, 2018
 #2
avatar+87639 
+1

Thanks, alan.....here's a non-Calculus qpproach

 

Let the product   xy   =  a    

 

And  xy will be maximized where this function is tangent to the ellipse...so we need to find  the "a" that makes this true

 

So   xy  = a   ⇒  y =  a/x

 

Subbing this into   x^2 + 2y^2  = 16   we get

 

x^2  + 2(a/x)^2  =  16       multiply through by   x^2

 

x^4  +  2a^2  =  16x^2

 

Let  v = x^2

 

v^2  -  16v  +  2a^2  =  0

 

We will have  a double real (positive) vaue for v whenever

256 - 8a^2  =  0

 

256 =  8a^2

 

a^2  =  32    ⇒  2a^2  =  64

 

a = √32  =  4√2

 

So    

 

v^2 - 16v  -  64 =  0

 

(v - 8)^2  =  0   ⇒ v =  8  ⇒  x^2  =  8  ⇒  x = ±√8  ⇒  x = ±2√2

 

And  y  =  (a/x)  =  (4√2 / ( ±2√2) )  =  ±2

 

So.....the product xy is maximized whenever  (x, y)  =  ( ±2√2, ±2)

 

And   xy  is  maximized  when  a  =  4√2    as this graph shows :

 

https://www.desmos.com/calculator/vmykhikdjb

 

 

cool cool cool

CPhill  Jan 23, 2018
edited by CPhill  Jan 23, 2018
 #3
avatar+19835 
+2

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

 

Formula: Ellipse equation
\(\begin{array}{|rcll|} \hline \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} &=& 1 \\ (xy)_{\text{max}} &=& \dfrac{ab}{2}\\ \hline \end{array}\)

 

Proof:

\(\begin{array}{|rcll|} \hline \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} &=& 1 \quad & | \quad \cdot x^2 \\\\ \dfrac{x^4}{a^2}+\dfrac{x^2y^2}{b^2} &=& x^2 \quad & | \quad \mathbf{z=xy} \\\\ \dfrac{x^4}{a^2}+\dfrac{\mathbf{z}^2}{b^2} &=& x^2 \\\\ \dfrac{x^4}{a^2}+\dfrac{\mathbf{z}^2}{b^2} &=& x^2 \quad & | \quad -\dfrac{x^4}{a^2} \\\\ \dfrac{\mathbf{z}^2}{b^2} &=& x^2 -\dfrac{x^4}{a^2} \quad & | \ \text{Differentiate each term with respect to x}\\\\ \dfrac{\mathbf{2z\ dz}}{b^2} &=& 2x\ dx -\dfrac{4x^3\ dx}{a^2} \quad & | \quad :2 \\\\ \dfrac{\mathbf{ z\ dz}}{b^2} &=& x\ dx -\dfrac{2x^3\ dx}{a^2} \quad & | \quad :\ dx \\\\ \dfrac{\mathbf{ z}}{b^2}\cdot\frac{\mathbf{dz}}{dx}\ &=& x-\dfrac{2x^3}{a^2} \quad & | \quad \text{set } \frac{ \mathbf{dz}}{dx} = 0 \Rightarrow \mathbf{ z}_{\text{max}} =(xy)_{\text{max}} \\\\ x-\dfrac{2x^3}{a^2} &=& 0 \\\\ x-\dfrac{2x^3}{a^2} &=& 0 \\\\ x\left(1- \dfrac{2x^2}{a^2} \right) &=& 0 \\ \hline \end{array}\)

 

1. \(x = 0\) no solution \(\Rightarrow xy = 0 \) no maximum

2. x = ?

\(\begin{array}{rcll} 1- \dfrac{2x^2}{a^2} &=& 0 \\\\ \dfrac{2x^2}{a^2} &=& 1 \\\\ x^2 &=& \dfrac{a^2}{2} \quad & | \quad \pm\sqrt{} \\\\ x &=& \pm \dfrac{a}{\sqrt{2}} \quad & | \quad \cdot \frac{\sqrt{2}}{\sqrt{2}} \\\\ \mathbf{ x } &\mathbf{=}& \mathbf{ \pm a\cdot \dfrac{\sqrt{2}}{2} } \end{array} \)

 

y = ?

\(\begin{array}{rcll} \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} &=& 1 \quad & | \quad \mathbf{ x } &\mathbf{=}& \mathbf{ \pm \dfrac{a}{\sqrt{2}} } \\\\ \dfrac{a^2}{2a^2}+\dfrac{y^2}{b^2} &=& 1 \\\\ \dfrac{1}{2}+\dfrac{y^2}{b^2} &=& 1 \quad & | \quad -\dfrac12 \\\\ \dfrac{y^2}{b^2} &=& 1 -\dfrac12 \\\\ \dfrac{y^2}{b^2} &=& \dfrac12 \\\\ y^2 &=& \dfrac{b^2}{2} \quad & | \quad \pm\sqrt{} \\\\ y &=& \pm \dfrac{b}{\sqrt{2}} \quad & | \quad \cdot \frac{\sqrt{2}}{\sqrt{2}} \\\\ \mathbf{ y } &\mathbf{=}& \mathbf{ \pm b\cdot \dfrac{\sqrt{2}}{2} } \end{array}\)

 

xy = ?

\(\begin{array}{|rcll|} \hline \mathbf{ z}_{\text{max}} =(xy)_{\text{max}} \\ (xy)_{\text{max}} &=& \mathbf{ \left(\pm a\cdot \dfrac{\sqrt{2}}{2} \right) } \cdot \mathbf{ \left( \pm b\cdot \dfrac{\sqrt{2}}{2} \right) } \quad & | \quad xy > 0! \\\\ &=& a\cdot \dfrac{\sqrt{2}}{2}\cdot b\cdot \dfrac{\sqrt{2}}{2} \\\\ &=& ab\cdot \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{2}}{2} \\\\ \mathbf{(xy)_{\text{max}}} &\mathbf{=}& \mathbf{ \dfrac{ab}{2} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x^2 + 2y^2 &=& 16 \quad & | \quad : 16 \\\\ \dfrac{x^2}{16} + \dfrac{2y^2}{16} &=& 1 \\\\ \dfrac{x^2}{16} + \dfrac{ y^2}{8} &=& 1 \\\\ \dfrac{x^2}{\mathbf{4}^2} + \dfrac{ y^2}{(\mathbf{\sqrt{8}})^2} &=& 1 \\\\ \text{so } a = 4 \text{ and } b = \sqrt{8} \\\\ \mathbf{(xy)_{\text{max}}} &\mathbf{=}& \mathbf{ \dfrac{ab}{2} } \\\\ &=& \dfrac{4 \sqrt{8}}{2} \\\\ &=& 2 \sqrt{8} \\ &=& 2 \sqrt{2\cdot4 } \\ &=& 2\cdot2 \sqrt{2} \\ \mathbf{(xy)_{\text{max}}} &\mathbf{=}& \mathbf{4 \sqrt{2}} \\ \hline \end{array} \)

 

\(\begin{array}{rcll} Point_1 &=& ( a\cdot \dfrac{\sqrt{2}}{2},~ b\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( 4\cdot \dfrac{\sqrt{2}}{2},~ \sqrt{8}\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( 2\cdot \sqrt{2} ,~ \dfrac{\sqrt{16}}{2} ) \\ &=& ( 2\cdot \sqrt{2} ,~ \dfrac{4}{2} ) \\ &=& ( 2\cdot \sqrt{2} ,~ 2 ) \\ \end{array} \begin{array}{rcll} Point_2 &=& ( -a\cdot \dfrac{\sqrt{2}}{2},~ -b\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( -4\cdot \dfrac{\sqrt{2}}{2}, -\sqrt{8}\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( -2\cdot \sqrt{2} ,~ -\dfrac{\sqrt{16}}{2} ) \\ &=& ( -2\cdot \sqrt{2} ,~ -\dfrac{4}{2} ) \\ &=& ( -2\cdot \sqrt{2} ,~ -2 ) \\ \end{array} \)

 

laugh

heureka  Jan 24, 2018
edited by heureka  Jan 24, 2018

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