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Dear-mr-w-i-want-discuss-for-equation-find-minimum-and-maximum-value-of-xy-2-with-constraint-x-2-y-2-6-my-way-short-cut-x-2-y-2-3-max-3-2-2-5-min-3-2-1




Question Number 86343 by jagoll last updated on 28/Mar/20
Dear mr w. i want   discuss for equation  find minimum and maximum value  of xy +2 with constraint  x^2 +y^(2 ) = 6.  my way (short cut)  ⇒ x^2  = y^2  = 3   { ((max = ((√3))^2 +2 = 5)),((min = −((√3))^2  +1 = −1)) :}  it correct?
$$\mathrm{Dear}\:\mathrm{mr}\:\mathrm{w}.\:\mathrm{i}\:\mathrm{want}\: \\ $$$$\mathrm{discuss}\:\mathrm{for}\:\mathrm{equation} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{xy}\:+\mathrm{2}\:\mathrm{with}\:\mathrm{constraint} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}\:} =\:\mathrm{6}. \\ $$$$\mathrm{my}\:\mathrm{way}\:\left(\mathrm{short}\:\mathrm{cut}\right) \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{3} \\ $$$$\begin{cases}{\mathrm{max}\:=\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}\:=\:\mathrm{5}}\\{\mathrm{min}\:=\:−\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:+\mathrm{1}\:=\:−\mathrm{1}}\end{cases} \\ $$$$\mathrm{it}\:\mathrm{correct}? \\ $$
Commented by mr W last updated on 28/Mar/20
this is correct, because both in xy+2  and in x^2 +y^2 =6 there is the symmetry  for x and y, i.e. x and y are equally  valued.
$${this}\:{is}\:{correct},\:{because}\:{both}\:{in}\:{xy}+\mathrm{2} \\ $$$${and}\:{in}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{6}\:{there}\:{is}\:{the}\:{symmetry} \\ $$$${for}\:{x}\:{and}\:{y},\:{i}.{e}.\:{x}\:{and}\:{y}\:{are}\:{equally} \\ $$$${valued}. \\ $$
Commented by jagoll last updated on 28/Mar/20
thank you very much mister
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{mister} \\ $$

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