Question

Show that among all positive numbers x and y with x² + y² =r², the sum x+y is largest when x=y=r \[\sqrt{2}\] .

Answer 1

\[\text { Here }, \]

\[ x^2 + y^2 = r^2 \]

\[ \Rightarrow y = \sqrt{r^2 - x^2} ................ \left( 1 \right)\]

\[\text { Now, }\]

\[Z = x + y\]

\[ \Rightarrow Z = x + \sqrt{r^2 - x^2} .............\left[ \text { From eq. } \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dZ}{dx} = 1 + \frac{\left( - 2x \right)}{2\sqrt{r^2 - x^2}}\]

\[\text { For maximum or minimum values of Z, we must have }\]

\[\frac{dZ}{dx} = 0\]

\[ \Rightarrow 1 + \frac{\left( - 2x \right)}{2\sqrt{r^2 - x^2}} = 0\]

\[ \Rightarrow 2x = 2\sqrt{r^2 - x^2}\]

\[ \Rightarrow x = \sqrt{r^2 - x^2}\]

\[\text { Squaring both the sides, we get }\]

\[ \Rightarrow x^2 = r^2 - x^2 \]

\[ \Rightarrow 2 x^2 = r^2 \]

\[ \Rightarrow x = \frac{r}{\sqrt{2}}\]

\[\text { Substituting the value of x in eq. } \left( 1 \right), \text { we get }\]

\[y = \sqrt{r^2 - x^2}\]

\[ \Rightarrow y = \sqrt{r^2 - \left( \frac{r}{\sqrt{2}} \right)^2}\]

\[ \Rightarrow y = \frac{r}{\sqrt{2}}\]

\[\frac{d^2 z}{d x^2} = \frac{- \sqrt{r^2 - x^2} + \frac{x\left( - x \right)}{\sqrt{r^2 - x^2}}}{r^2 - x^2}\]

\[ \Rightarrow \frac{d^2 z}{d x^2} = \frac{- r^2 + x^2 - x^2}{\left( r^2 - x^2 \right)^\frac{3}{2}}\]

\[ \Rightarrow \frac{d^2 z}{d x^2} = \frac{- r^2}{r^3} \times 2\sqrt{2}\]

\[ \Rightarrow \frac{d^2 z}{d x^2} = \frac{- 2\sqrt{2}}{r} < 0\]

\[\text { So, z = x + y is maximum when x = y } = \frac{r}{\sqrt{2}} . \]

\[\text { Hence proved } .\]