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 = ?
\(\begin{array}{|rcll|} \hline \mathbf{ z}_{\text{max}} =(xy)_{\text{max}} \\ (xy)_{\text{max}} &=& \mathbf{ \left(\pm a\cdot \dfrac{\sqrt{2}}{2} \right) } \cdot \mathbf{ \left( \pm b\cdot \dfrac{\sqrt{2}}{2} \right) } \quad & | \quad xy > 0! \\\\ &=& a\cdot \dfrac{\sqrt{2}}{2}\cdot b\cdot \dfrac{\sqrt{2}}{2} \\\\ &=& ab\cdot \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{2}}{2} \\\\ \mathbf{(xy)_{\text{max}}} &\mathbf{=}& \mathbf{ \dfrac{ab}{2} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline x^2 + 2y^2 &=& 16 \quad & | \quad : 16 \\\\ \dfrac{x^2}{16} + \dfrac{2y^2}{16} &=& 1 \\\\ \dfrac{x^2}{16} + \dfrac{ y^2}{8} &=& 1 \\\\ \dfrac{x^2}{\mathbf{4}^2} + \dfrac{ y^2}{(\mathbf{\sqrt{8}})^2} &=& 1 \\\\ \text{so } a = 4 \text{ and } b = \sqrt{8} \\\\ \mathbf{(xy)_{\text{max}}} &\mathbf{=}& \mathbf{ \dfrac{ab}{2} } \\\\ &=& \dfrac{4 \sqrt{8}}{2} \\\\ &=& 2 \sqrt{8} \\ &=& 2 \sqrt{2\cdot4 } \\ &=& 2\cdot2 \sqrt{2} \\ \mathbf{(xy)_{\text{max}}} &\mathbf{=}& \mathbf{4 \sqrt{2}} \\ \hline \end{array} \)
\(\begin{array}{rcll} Point_1 &=& ( a\cdot \dfrac{\sqrt{2}}{2},~ b\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( 4\cdot \dfrac{\sqrt{2}}{2},~ \sqrt{8}\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( 2\cdot \sqrt{2} ,~ \dfrac{\sqrt{16}}{2} ) \\ &=& ( 2\cdot \sqrt{2} ,~ \dfrac{4}{2} ) \\ &=& ( 2\cdot \sqrt{2} ,~ 2 ) \\ \end{array} \begin{array}{rcll} Point_2 &=& ( -a\cdot \dfrac{\sqrt{2}}{2},~ -b\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( -4\cdot \dfrac{\sqrt{2}}{2}, -\sqrt{8}\cdot \dfrac{\sqrt{2}}{2} ) \\ &=& ( -2\cdot \sqrt{2} ,~ -\dfrac{\sqrt{16}}{2} ) \\ &=& ( -2\cdot \sqrt{2} ,~ -\dfrac{4}{2} ) \\ &=& ( -2\cdot \sqrt{2} ,~ -2 ) \\ \end{array} \)
+9466 
+2440 
