The given function is $$f(x,y)= 5x +7y$$ and the constraint is $$g(x,y)=x^2y-10=0$$ First I took the gradient of $f$ and $g$ and got $$f(x,y)=\left<5,7 \right>$$ $$\lambda g(x,y)=\left< 2xy, x^2\right>$$ Then I have two equations $$5=2xy\lambda$$ $$7=x^2 \lambda$$ Solving for $\lambda$ I get $$\lambda = \frac{5xy}{2}$$ Plugging this into one of the two equations I get $$7 =x^2(\frac{5xy}{2})$$ Which then works out to $$x=\sqrt{\frac{7}{\lambda}}$$ Then solving for $y$ I get $$y= \frac{5\sqrt{7}\sqrt{\lambda}}{14\lambda}$$ And then $\lambda$ then works out to be $$\lambda = \frac{\sqrt[3]{7}*2^{2/3}}{4}$$ Which is approximately $0.7591$ This doesn't feel right to me what with all the weird terms and powers. Did I do it right?

  • 1
    $\begingroup$ Yes your working is correct. I get the same but you can find $x$ and $y$ without calculating the value of $\lambda$. $\endgroup$
    – Math Lover
    Dec 10, 2020 at 16:33
  • $\begingroup$ I'm not sure if i did the rest of it right. I solved for $x$ and $y$ using $\lambda$ but the answers I got are a bit strange. $x=\sqrt{\frac{7}{\sqrt[3]{\frac{175}{400}}}}$ and $y=\frac{sqrt[3]{25} * 10^{2/3}}{14}$ $\endgroup$ Dec 10, 2020 at 18:05
  • $\begingroup$ I think it is correct. Though it looks more complicated as is, you should be able to simplify it. I just posted how I did without really finding $\lambda$ and then plugging back in, that way you can directly get $x$ and $y$ and you get a simpler expression to begin with. $\endgroup$
    – Math Lover
    Dec 10, 2020 at 19:01
  • $\begingroup$ I'ld solve this problem using Lagrange only if I'm getting paid for it: in less that 15 second I got $x^3=28$. $\endgroup$ Dec 10, 2020 at 19:32
  • $\begingroup$ @MichaelHoppe completely agree on that. $\endgroup$
    – Math Lover
    Dec 10, 2020 at 19:51

1 Answer 1


$x^2λ = 7 \implies \lambda = \frac{7}{x^2}$ ...(i)

$2xyλ = 5$ ...(ii)

Plugging value of $\lambda$ from (i) into (ii)

$\displaystyle \frac{14y}{x} = 5 \implies x = \frac{14y}{5}$ ...(iii)

Now plugging $x$ into our constraint,

$x^2y = 10 \implies y^3 = \frac{250}{196}$

$\displaystyle y = (\frac{250}{196})^{1/3} \, \approx 1.0845$

Using (iii), $x \displaystyle = \frac{14}{5} (\frac{250}{196})^{1/3} = (28)^{1/3} \, \approx 3.0366$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .