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the equations : $$\left\{\begin{array}{l}{x+6 y+4 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$ I've done many transformations,but I still can't get the answer.

the last I did this: $$\lambda=\left(6 \frac{x}{y}+2\right)=\left(\frac{1}{4}+\frac{3}{2} \frac{y}{x}\right),y=tx \\=\left(6 \cdot \frac{1}{t}+2\right)=\left(\frac{1}{4}+\frac{3}{2} t\right)\\t^{2}-\frac{7}{6} t-4=0$$

the quadratic formula's roots have square number , but the answer doesn't have,so I think I might be wrong.

I don't know how to solve it

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Consider the first two equations, under this slightly different form:$$\left\{\begin{array}{l}(1+4\lambda)x+6y=0\\6x+(2+\lambda)y=0.\end{array}\right.$$Suppose that this homogeneous system has exatly one solution; then this solution is $(x,y)=(0,0)$, which is not a solution of the third one.

But\begin{align}\text{The system has more than one solution}&\iff\det\begin{bmatrix}1+4\lambda&6\\6&2+\lambda\end{bmatrix}=0\\&\iff\lambda=-\frac{17}4\vee\lambda=2.\end{align}So, deal only with the cases $\lambda=-\frac{17}4$ and $\lambda=2$. Can you do that?

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Well, the first equation gives:

$$x+6y+4\lambda x=6y+x(1+4\lambda)=0\space\Longleftrightarrow\space x=-\frac{6y}{1+4\lambda}\tag1$$

Using the second equation, we get:

$$6x+2y+\lambda y=0\space\Longleftrightarrow\space x=-\frac{2y+\lambda y}{6}\tag2$$

So, we get:

$$-\frac{2y+\lambda y}{6}=-\frac{6y}{1+4\lambda}\space\Longleftrightarrow\space2+\lambda=\frac{36}{1+4\lambda}\space\Longleftrightarrow\space\lambda=-\frac{17}{2}\space\vee\lambda=2\tag3$$

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    $\begingroup$ At the moment you have cancelled $y$ without establishing that $y\neq 0$ $\endgroup$ – Mark Bennet Aug 16 at 20:39
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From the first equation we get $$\frac{y}{x}=-\frac{1+4\lambda}{6}$$ and from the second one $$\frac{y}{x}=-\frac{6}{2+\lambda}$$ So you will get $$\frac{1+4\lambda}{6}=\frac{6}{2+\lambda}$$ Can you finish?

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    $\begingroup$ As stated you have not established that you are not dividing by zero. $x=y=0$ satisfies the first two equations, but you have not excluded it. $\endgroup$ – Mark Bennet Aug 16 at 20:40
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Restoring the objective function from the constraint (third) equation and partial derivatives: $$Optimize \ \ z(x,y)=\frac12x^2+6xy+y^2 \ \ s.t. \ \ 4x^2+y^2-25=0.$$ If you take partial derivatives now: $$\left\{\begin{array}{l}{x+6 y+\color{red}8 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$ Note that if $x=0$, then from the first equation $y=0$, but it contradicts the last equation. So, $x\ne 0$. Similarly, if $y=0$, then from the second equation $x=0$, but it contradicts the last equation. So, $y\ne 0$ too. Thus: $$\begin{cases}y=\frac{-(1+8\lambda)x}{6}\\ 6x+(2+\lambda)\frac{-(1+8\lambda)x}{6}=0 \Rightarrow x(36-(2+\lambda)(1+8\lambda))=0 \Rightarrow 36-(2+\lambda)(1+8\lambda)=0\end{cases} \Rightarrow \\ 8\lambda^2+17\lambda-34=0 \Rightarrow \lambda_{1,2}=\frac{-17\pm 9\sqrt{17}}{16}. $$ Can you do the rest?

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