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Question-191454




Question Number 191454 by Mingma last updated on 24/Apr/23
Commented by mr W last updated on 24/Apr/23
xyz=((646)/(25))
$${xyz}=\frac{\mathrm{646}}{\mathrm{25}} \\ $$
Commented by Mingma last updated on 24/Apr/23
Good!
Answered by mr W last updated on 24/Apr/23
(√(4^2 −x^2 ))+(√(5^2 −y^2 ))+(√(6^2 −z^2 ))  ≥(√((4+5+6)^2 −(x+y+z)^2 ))  =(√(15^2 −9^2 ))  =12  equality holds only if   (x/4)=(y/5)=(z/6), i.e. x=4k, y=5k, z=6k  (4+5+6)k=9, k=(9/(15))=(3/5)  ⇒xyz=4×5×6×k^3 =120×((3/5))^3 =((646)/(25)) ✓
$$\sqrt{\mathrm{4}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\sqrt{\mathrm{5}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\sqrt{\mathrm{6}^{\mathrm{2}} −{z}^{\mathrm{2}} } \\ $$$$\geqslant\sqrt{\left(\mathrm{4}+\mathrm{5}+\mathrm{6}\right)^{\mathrm{2}} −\left({x}+{y}+{z}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{15}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} } \\ $$$$=\mathrm{12} \\ $$$${equality}\:{holds}\:{only}\:{if}\: \\ $$$$\frac{{x}}{\mathrm{4}}=\frac{{y}}{\mathrm{5}}=\frac{{z}}{\mathrm{6}},\:{i}.{e}.\:{x}=\mathrm{4}{k},\:{y}=\mathrm{5}{k},\:{z}=\mathrm{6}{k} \\ $$$$\left(\mathrm{4}+\mathrm{5}+\mathrm{6}\right){k}=\mathrm{9},\:{k}=\frac{\mathrm{9}}{\mathrm{15}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow{xyz}=\mathrm{4}×\mathrm{5}×\mathrm{6}×{k}^{\mathrm{3}} =\mathrm{120}×\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{3}} =\frac{\mathrm{646}}{\mathrm{25}}\:\checkmark \\ $$
Commented by Mingma last updated on 24/Apr/23
Very nice, sir!

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