minimum-value-f-x-y-x-2-y-2-with-constrain-g-x-y-x-2-y-16-

Question Number 101258 by bemath last updated on 01/Jul/20

minimum value f (x, y) = x^{2} + y^{2}

with constrain g (x, y) = x^{2} y - 16

Commented by john santu last updated on 01/Jul/20

f (x, y, λ) = x^{2} + y^{2} + λ (x^{2} y - 16)

\frac{\partial f}{\partial x} = 2 x + λ (2 xy) = 0 \to λ = \frac{- 2 x}{2 xy} = - \frac{1}{y}

\frac{\partial f}{\partial y} = 2 y + λ x^{2} = 0 \to λ = - \frac{2 y}{x^{2}}

\Leftrightarrow - \frac{1}{y} = \frac{- 2 y}{x^{2}} \Rightarrow y^{2} = \frac{1}{2} x^{2}

\frac{\partial f}{\partial λ} = x^{2} y - 16 = 0 \to x^{2} = \frac{16}{y}

we get y^{2} = \frac{1}{2} (\frac{16}{y}) \Rightarrow y^{3} = 8

y = 2 \land x = \pm 2 \sqrt{2}

{\begin{cases} for (2 \sqrt{2}, 2) \Rightarrow f (2 \sqrt{2}, 2) = 12 \\ for (- 2 \sqrt{2}, 2) \Rightarrow f (- 2 \sqrt{2}, 2) = 12 \end{cases}

minimum value is 12 ★

Commented by bramlex last updated on 02/Jul/20

♢ ♣ ⊸

Answered by mr W last updated on 01/Jul/20

y = \frac{16}{x^{2}}

f (x) = x^{2} + {(\frac{16}{x^{2}})}^{2}

\frac{d f}{d x} = 2 x + 2 (\frac{16}{x^{2}}) (- \frac{2 \times 16}{x^{3}}) = 0

x - \frac{2^{9}}{x^{5}} = 0

\Rightarrow x^{2} = 8

f_{m i n} = 8 + {(\frac{16}{8})}^{2} = 12

Answered by 1549442205 last updated on 07/Jul/20

f (x, y) = h (y) = \frac{16}{y} + y^{2} \Rightarrow h^{'} (y) = - \frac{16}{y^{2}} + 2 y = 0

\Leftrightarrow {2 y}^{3} - 16 = 0 \Leftrightarrow y = 2, h^{″} (y) = \frac{32}{y^{3}} + 2

h^{″} (2) = 6 > 0 \Rightarrow h_{\min} (y) = h (2) = 12

\Rightarrow f {(x, y)}_{\min} = h_{\min} (y) = 12 when y = 2

x^{2} = \frac{16}{2} = 8 \Leftrightarrow x = \pm \sqrt{2}

Thus, f_{\min} (x, y) = 12 when (x; y) \in {(- 2 \sqrt{2}; 2); (2 \sqrt{2}; 2)}

other way :

f (x, y) = g (x) = x^{2} + {(\frac{16}{x^{2}})}^{2} = x^{2} + \frac{256}{x^{4}}

= \frac{x^{2}}{2} + \frac{x^{2}}{2} + \frac{256}{x^{4}} ⩾ 3^{3} \sqrt{\frac{x^{2}}{2} . \frac{x^{2}}{2} . \frac{256}{x^{4}}} =

= 3^{3} \sqrt{64} = 3 \times 4 = 12 .

Equality ocurrs if and only if

{\begin{cases} \frac{x^{2}}{2} = \frac{256}{x^{4}} \\ x^{2} y = 16 \end{cases} \Leftrightarrow {\begin{cases} x^{2} = 8 \\ y = 2 \end{cases} \Leftrightarrow (x, y) \in {(- 2 \sqrt{2}; 2); (2 \sqrt{2}; 2)}

Thus, f_{\min} (x, y) = g_{\min} (x) = 12 when

(x, y) \in {(- 2 \sqrt{2}; 2); (2 \sqrt{2}; 2)}