How do I use the Lagrange multiplier in this problem? Suppose I have this problem:
max     f(x,y)
s.t    2x + 4y = 3

If the function $f(x,y) = x^{0.7}y^{0.3}$ then how can I use Lagrange multiplier to find the maximum input choices? I found out that $\nabla$$g(x,y) = 2i + 4j$ and $\nabla$$f(x,y) = 0.7(y/x)^{0.3} + 0.3(x/y)^{0.7}$. So I know that $\nabla$f = λ$\nabla$g but how do I use these numbers to find the maximum input choices? 
 A: Consider $$F=x^{7/10} y^{3/10}+\lambda  (2 x+4 y-3)$$ compute the derivatives and set them equal to zero $$F'_x=2 \lambda +\frac{7 y^{3/10}}{10 x^{3/10}}=0$$ $$F'_y=4 \lambda +\frac{3 x^{7/10}}{10 y^{7/10}}=0$$ $$F'_\lambda=2 x+4 y-3=0$$ From $F'_x=0$, you can extract $$\lambda=-\frac{7 y^{3/10}}{20 x^{3/10}}$$ Replacing $\lambda$ in $F'_y$, you get $$F'_y=\frac{3 x^{7/10}}{10 y^{7/10}}-\frac{7 y^{3/10}}{5 x^{3/10}}=\frac{3 x-14 y}{10 x^{3/10} y^{7/10}}=0$$ which makes $$3x-14y=0$$ Combine this with $F'_\lambda$ : two linear equation for two unknowns $x,y$. Solve and compute the value of $x^{7/10} y^{3/10}$.
At first glance, I thought that the problem is simple because the sum of the exponents is equal to $1$. Effectively, making it more general such as $$F=x^{a} y^{1-a}+\lambda  (2 x+4 y-3)$$ just doing the same, you should arrive to $$(1-a)x-2ay=0$$
But, the problem stays simple if set as $$F=x^{a} y^{b}+\lambda  (2 x+4 y-3)$$ Repeat the same steps to arrive to $b x-2ay=0$.
I am sure than you can take it from here.
A: during the step
Replacing λ in F'y, 
you divided the denominator by 4 because of 4λ. Can you actually do this? It seems like it would change the value.
