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Question Number 141135 by bobhans last updated on 16/May/21
Find the minimum of ((12)/x) + ((18)/y) + xy for all positive number x & y .
$$\:\:\:\:\:\:\:{Find}\:{the}\:{minimum}\:{of}\: \\ $$$$\:\:\:\:\:\:\frac{\mathrm{12}}{{x}}\:+\:\frac{\mathrm{18}}{{y}}\:+\:{xy}\:{for}\:{all}\: \\ $$$$\:\:\:\:\:\:{positive}\:{number}\:{x}\:\&\:{y}\:. \\ $$
Answered by mitica last updated on 16/May/21
((12)/x)+((18)/y)+xy≥3((((12)/x)∙((18)/y)∙xy))^(1/3) =18 = for ((12)/x)=((18)/y)=xy⇔x=2,y=3 ⇒min=18
$$\frac{\mathrm{12}}{{x}}+\frac{\mathrm{18}}{{y}}+{xy}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{12}}{{x}}\centerdot\frac{\mathrm{18}}{{y}}\centerdot{xy}}=\mathrm{18} \\ $$$$=\:{for}\:\frac{\mathrm{12}}{{x}}=\frac{\mathrm{18}}{{y}}={xy}\Leftrightarrow{x}=\mathrm{2},{y}=\mathrm{3} \\ $$$$\Rightarrow{min}=\mathrm{18} \\ $$
Commented by mitica last updated on 16/May/21
x=2,y=3
$${x}=\mathrm{2},{y}=\mathrm{3} \\ $$
Commented by iloveisrael last updated on 16/May/21
x = ((4/3))^(1/3) ; y = ((9/2))^(1/3) ⇒((12)/( ((4/3))^(1/3) )) + ((18)/( ((9/2))^(1/3) ))+(((36)/6))^(1/3) = 23.62 how you claim 23.62 = 18 ??
$${x}\:=\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}}{\mathrm{3}}}\:;\:{y}\:=\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{12}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}}{\mathrm{3}}}}\:+\:\frac{\mathrm{18}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{36}}{\mathrm{6}}}\:\:=\:\mathrm{23}.\mathrm{62} \\ $$$${how}\:{you}\:{claim}\:\mathrm{23}.\mathrm{62}\:=\:\mathrm{18}\:?? \\ $$
Answered by iloveisrael last updated on 16/May/21
We know that the three terms in the sum have fixed product 12×18 = 216 . The sum is therefore a minimum if we can find x &y satisfy ((12)/x) = ((18)/y) = xy . Since the product is 6^3 each of ((12)/x) , ((18)/y) & xy must equal 6 , hence the required minimum is 6+6+6 = 18 , occurring iff x=2 &y = 3.
$$\:{We}\:{know}\:{that}\:{the}\:{three}\:{terms} \\ $$$$\:{in}\:{the}\:{sum}\:{have}\:{fixed}\:{product} \\ $$$$\:\mathrm{12}×\mathrm{18}\:=\:\mathrm{216}\:.\:{The}\:{sum}\:{is}\:{therefore} \\ $$$$\:{a}\:{minimum}\:{if}\:{we}\:{can}\:{find}\:{x}\:\&{y} \\ $$$${satisfy}\:\frac{\mathrm{12}}{{x}}\:=\:\frac{\mathrm{18}}{{y}}\:=\:{xy}\:.\:{Since}\:{the} \\ $$$${product}\:{is}\:\mathrm{6}^{\mathrm{3}} \:{each}\:{of}\:\frac{\mathrm{12}}{{x}}\:,\:\frac{\mathrm{18}}{{y}}\:\&\:{xy}\: \\ $$$${must}\:{equal}\:\mathrm{6}\:,\:{hence}\:{the}\:{required}\: \\ $$$${minimum}\:{is}\:\mathrm{6}+\mathrm{6}+\mathrm{6}\:=\:\mathrm{18}\:, \\ $$$${occurring}\:{iff}\:{x}=\mathrm{2}\:\&{y}\:=\:\mathrm{3}.\: \\ $$
Answered by EDWIN88 last updated on 16/May/21
z = f(x,y) = 12x^(−1) +18y^(−1) +xy ⇒z_x = −((12)/x^2 ) +y = 0 , y = ((12)/x^2 ) …(i) ⇒z_y = −((18)/y^2 ) +x = 0, x = ((18)/y^2 ) …(ii) ⇒ ((12)/(18)) = ((x^2 y)/(xy^2 )) ; (2/3) = (x/y) or y = ((3x)/2) ⇒((3x)/2) = ((12)/x^2 ) ⇒ x = (8)^(1/3) = 2 & y = 3 minimum z = f(2,3) = 18 ⋇
$$\:\mathrm{z}\:=\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{12x}^{−\mathrm{1}} +\mathrm{18y}^{−\mathrm{1}} +\mathrm{xy} \\ $$$$\:\Rightarrow\mathrm{z}_{\mathrm{x}} \:=\:−\frac{\mathrm{12}}{\mathrm{x}^{\mathrm{2}} }\:+\mathrm{y}\:=\:\mathrm{0}\:,\:\mathrm{y}\:=\:\frac{\mathrm{12}}{\mathrm{x}^{\mathrm{2}} }\:\ldots\left(\mathrm{i}\right) \\ $$$$\Rightarrow\mathrm{z}_{\mathrm{y}} \:=\:−\frac{\mathrm{18}}{\mathrm{y}^{\mathrm{2}} }\:+\mathrm{x}\:=\:\mathrm{0},\:\mathrm{x}\:=\:\frac{\mathrm{18}}{\mathrm{y}^{\mathrm{2}} }\:\ldots\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{12}}{\mathrm{18}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{y}}{\mathrm{xy}^{\mathrm{2}} }\:;\:\frac{\mathrm{2}}{\mathrm{3}}\:=\:\frac{\mathrm{x}}{\mathrm{y}}\:\mathrm{or}\:\mathrm{y}\:=\:\frac{\mathrm{3x}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{3x}}{\mathrm{2}}\:=\:\frac{\mathrm{12}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\:\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{8}}\:=\:\mathrm{2}\:\&\:\mathrm{y}\:=\:\mathrm{3}\: \\ $$$$\mathrm{minimum}\:\mathrm{z}\:=\:\mathrm{f}\left(\mathrm{2},\mathrm{3}\right)\:=\:\mathrm{18}\:\divideontimes\: \\ $$

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