Use Lagrange multipliers to find the exact minimum value

Question

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?

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