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# Show that Among All Positive Numbers X and Y with X2 + Y2 =R2, the Sum X+Y is Largest When X=Y=R √ 2 . - Mathematics

#### Question

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

#### Solution

$\text { Here },$

$x^2 + y^2 = r^2$

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

$\text { Now, }$

$Z = x + y$

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

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

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

$\frac{dZ}{dx} = 0$

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

$\Rightarrow 2x = 2\sqrt{r^2 - x^2}$

$\Rightarrow x = \sqrt{r^2 - x^2}$

$\text { Squaring both the sides, we get }$

$\Rightarrow x^2 = r^2 - x^2$

$\Rightarrow 2 x^2 = r^2$

$\Rightarrow x = \frac{r}{\sqrt{2}}$

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

$y = \sqrt{r^2 - x^2}$

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

$\Rightarrow y = \frac{r}{\sqrt{2}}$

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

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

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

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

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

$\text { Hence proved } .$

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