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How would I solve this system of equation? I guess Lagrange Multiplier is important when solving this.

$$\begin{align}&y^2=5\lambda\\ &2xy=3\lambda\\ &3x+4y=11 \end{align}$$

I have tried to solve this for one hour now but some reason I still don't seem to get in the right path.

Could someone please help me with this one?

EDIT: After shuffling all the numbers I get Then $ x= \frac{33}{49}$ and $y=\frac{110}{49}$.

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  • $\begingroup$ Following the given hint we have y=x10/3 for sure. From here we can easily find x and y. Therefore your solution seems correct. $\endgroup$
    – user
    Sep 26, 2019 at 13:49

5 Answers 5

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You can “eliminate $\lambda$: multiply the first equation by $3$ and the second equation by $5$; then subtract the second from the first, getting $$ 3y^2-10xy=0 $$ Thus either $y=0$ or $y=10x/3$. Can you finish?

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You can try to divided the first two equation. Then $\frac{y}{2x}=\frac53$. Can you finish it?

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  • $\begingroup$ As a final solution I get $ x= \frac{33}{49}$ and $y=\frac{110}{49}$. Do you think this is correct? $\endgroup$
    – MiMaKo
    Sep 26, 2019 at 13:45
  • $\begingroup$ @MiMaKo No, you're forgetting a solution. $\endgroup$
    – egreg
    Sep 26, 2019 at 22:12
  • $\begingroup$ @egreg Okay, that's cryptic. Do you mean lambda is the correct answer? If so, then it is $\lambda=\frac{2420}{2401}$ $\endgroup$
    – MiMaKo
    Sep 26, 2019 at 22:27
  • $\begingroup$ @MiMaKo One cannot divide by zero, which is the problem with this answer. If you do it correctly, you get either $y=0$ (hence $x=11/3$ and $\lambda=0$) or $y=10x/3$ (so $x=33/49$, $y=110/49$ and $\lambda=2420/2401$). $\endgroup$
    – egreg
    Sep 26, 2019 at 22:38
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Put $ 3x + 4y = 11 $ in terms of $ x $ and substitute into $ 2xy = 3L $. Then substitute the resulting equation into $ y^2 = 5L $ to arrive at

$$ 22y - 13y^2 = 0 $$

From there you can use the quadratic formula to find $y$.

Once you have y, substitute the value found in the equations and solve for $x$ and $L$

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We have

$$y^2=5\lambda,\; 2xy=3\lambda \implies3y^2=10xy \implies y(3y-10x)=0$$

then consider two cases

  • $y=0$ not acceptable
  • $y\neq 0 \implies y=\frac{10}3 x\implies 3x+4\frac{10}3 x=11 \implies x=\cdots$
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If $x=0$, we get by the third equation that $y=11/4$ and then the value of $\lambda$ follows by the first equation. Then we assume $x\neq 0$ and by the second equation $y=(3\lambda)/(2x)$. Using this, we get a system of two equations in $x$ and $\lambda$:

$9\lambda^2=(5\lambda)(4x^2)$ and $3x^2+6\lambda=11x$.

Now, again, if $\lambda=0$ we get the value of $x$. Then, again, we assume $\lambda\neq 0$, so that, by $9\lambda^2=(5\lambda)(4x^2)$, we get that $\lambda=(20/9)x^2$ and we can determine the value of $x$ using $3x^2+6\lambda=11x$.

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