I am fairly new to Lagrange multipliers. Can someone please show me how to maximize the following function: \begin{align} f(x,y)=240\sqrt{x}+y \end{align} Subject to: \begin{align} 30x+y=720 \end{align}
Here is what I have done so far, is this on the right track? are there any errors? \begin{align} g(x,y,λ)=240\sqrt{x}+y+λ(720-30x+y) \end{align} \begin{align} {\frac{df}{dx}} = 120{x^{-1/2}}+λ=0 \end{align} \begin{align} {\frac{df}{dy}} = 2+λ=0 \end{align} \begin{align} {\frac{df}{dx}} = 720-30x+y=0 \end{align} I'm having trouble continuing from here.
First of all your transposing of the constraint is not right.
$720=30x+y\Rightarrow 720-30x-y=0$. Therefore the lagrange function is
$g(x,y,λ)=240\sqrt{x}+y+λ(720-30x-y)$
Then the partial derivatives come next. Pay attention on the corresponding symbols.
${\frac{\partial f}{\partial x}} = 120{x^{-1/2}}-30\lambda=0$
${\frac{\partial f}{\partial y}} = 1-λ=0$
${\frac{\partial f}{\partial \lambda}} = 720-30x-y=0$
Are the partial derivatives comprehensible ? You have some errors. From the second equation we can conclude that $\lambda=1$. This information can be used to caclulate the value of x. The first equation becomes
$120{x^{-1/2}}-30\cdot 1=0$
$120{x^{-1/2}}=30\quad \quad |\cdot x^{1/2}$
$120=30\cdot x^{1/2}$
$4=x^{1/2}\Rightarrow x^*=16$
The third equation can be used to calculate the optimal value of y.
$720-30\cdot 16-y=0 \Rightarrow 720-480=y\Rightarrow y^*=240$.
Finally you have to calculate $f(x^*,y^*)$.