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Question Number 96500 by bemath last updated on 02/Jun/20

$$\mathrm{If}xandyrealnumbersatisfy$$$${(x+5)}^2+{(\mathrm{y}-12)}^2=196,\mathrm{then}$$$$\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}{x}^2+y^{2}is$$

Answered by bobhans last updated on 02/Jun/20

$$\mathrm{Let}x+5=14\cos \theta ,\mathrm{y}-12=14\sin \theta$$$$x^2+y^2=365+28(12\sin \theta -5\cos \theta )$$$$=365+364\sin (\theta -\beta )\mathrm{where}\tan \beta =\frac{5}{12}$$$$\mathrm{so}{x}^2+y^{2}hastheminimumvalue1$$$$when\theta =\frac{3\pi }{2}+\arctan \left(\frac{5}{12}\right),\mathrm{i}.\mathrm{e}$$$$\Rightarrow x=\frac{5}{13}\mathrm{\&}\mathrm{y}=-\frac{12}{13}$$

Commented by bemath last updated on 02/Jun/20

$$\mathrm{thanks}$$

Commented by 1549442205 last updated on 02/Jun/20

$$\mathrm{I}\mathrm{add}\mathrm{a}\mathrm{question}:\mathrm{Which}\mathrm{is}\mathrm{the}\mathrm{greatest}$$$$\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{expression}{\mathrm{x}}^2+{\mathrm{y}}^2?$$

Commented by MJS last updated on 02/Jun/20

$$365+364=729$$

Commented by bobhans last updated on 02/Jun/20

$$\mathrm{yes}===$$

Commented by bobhans last updated on 02/Jun/20

$$\mathrm{when}\theta =\frac{\pi }{2}+\arctan \frac{5}{12}$$

Commented by 1549442205 last updated on 02/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{all}\mathrm{right}!\mathrm{it}\mathrm{is}\mathrm{as}\mathrm{following}:$$$$\mathrm{we}\mathrm{have}{\mathrm{x}}^2+10\mathrm{x}+25+{\mathrm{y}}^2-24\mathrm{y}+144=196$$$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=27-10\mathrm{x}+24\mathrm{y}\left(1\right).\mathrm{Applyin}$$$${\mathrm{Bunhiacopxky}}^{\prime }\mathrm{s}\mathrm{inequality}\mathrm{we}\mathrm{get}$$$${(-10\mathrm{x}+24\mathrm{y})}^2⩽({10}^2+{24}^2)({\mathrm{x}}^2+{\mathrm{y}}^2)$$$$=676({\mathrm{x}}^2+{\mathrm{y}}^2)\Rightarrow \mid -10\mathrm{x}+24\mathrm{y}\mid ⩽26\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\left(2\right)$$$$\mathrm{From}\left(1\right),\left(2\right)\mathrm{we}\mathrm{get}{\mathrm{x}}^2+{\mathrm{y}}^2⩽27$$$$+26\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\iff (\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}+1)(\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}-27)⩽0$$$$\Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}⩽27\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2⩽729.$$$$\mathrm{The}\mathrm{equality}\mathrm{occurs}\mathrm{if}\mathrm{and}\mathrm{only}$$$$\mathrm{if}\{\begin{array}{l}\frac{\mathrm{x}}{-10}=\frac{\mathrm{y}}{24}\\ {\mathrm{x}}^2+{\mathrm{y}}^2=729\end{array}\iff \{\begin{array}{l}\mathrm{x}=\frac{-135}{13}\\ \mathrm{y}=\frac{324}{13}\end{array}$$$$$$

Answered by john santu last updated on 02/Jun/20

$$\mathrm{i}\mathrm{have}\mathrm{a}\mathrm{method}\mathrm{in}\mathrm{short}$$$$\mathrm{if}{(\mathrm{x}-\mathrm{a})}^2+{(\mathrm{y}-\mathrm{b})}^2={\mathrm{r}}^2$$$$\mathrm{then}{\mathrm{x}}^2+{\mathrm{y}}^2\{\begin{array}{l}\max ={\{\mathrm{r}+\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\}}^2\\ \min ={\{\mathrm{r}-\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\}}^2\end{array}$$

Commented by bobhans last updated on 02/Jun/20

$$\mathrm{waw}====\mathrm{thanks}$$

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