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Question Number 79404 by jagoll last updated on 25/Jan/20

$$\mathrm{for}\mathrm{every}\mathrm{real}\mathrm{number}\mathrm{a},\mathrm{b}$$$$\mathrm{such}\mathrm{that}{\mathrm{a}}^2+{\mathrm{b}}^2-4\mathrm{a}-6\mathrm{b}=2.$$$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{and}$$$$\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{the}$$$$\mathrm{expression}$$$$\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2-8\mathrm{a}-10\mathrm{b}+41}?$$

Commented by john santu last updated on 25/Jan/20

$$letf=\sqrt{a^2+b^2-4a-6b-(4a+4b)+41}$$$$f=\sqrt{43-(4a+4b)}.Thatthe$$$$valueoffwillbemaximum$$$$if4a+4bisminimum.$$$$let:4x+4y=l,thisisthetangent$$$$ofcircle{(a-2)}^2+{(b-3)}^2=15$$$$d=\frac{\mid 8+12-l\mid }{4\sqrt{2}},d=r$$$$4\sqrt{30}=\mid 20-l\mid$$$$20-l=\pm 4\sqrt{30}\Rightarrow l=20\pm 4\sqrt{30}$$$$\left(i\right)l=20+4\sqrt{30}$$$$f=\sqrt{43-(20+4\sqrt{30})}=\sqrt{23-2\sqrt{120}}$$$$=\sqrt{15}-2\sqrt{2}$$$$\left(ii\right)l=20-4\sqrt{30}$$$$f=\sqrt{43-(20-4\sqrt{30})}=\sqrt{15}+2\sqrt{2}$$$$\therefore f_{max}=\sqrt{15}+2\sqrt{2}$$$$f_{min}=\sqrt{15}-2\sqrt{2}$$

Commented by jagoll last updated on 25/Jan/20

$$\mathrm{great}!\mathrm{sir}\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 25/Jan/20

$$great$$

Commented by john santu last updated on 25/Jan/20

$$thankssir$$

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