Find the maximum and minimum values of f(x,y)=3x+y on the ellipse x^2+36y^2=1?

Question

dillonmerv · Answer

I found the value of λ to be ± (5√(13))/12. Is this correct? What do I do next? Can you please show the steps?

ganeshie8 · Answer

$3=2x\lambda$
$1=72y\lambda$
$x^2+36y^2 = 1$

Dividing first two equations gives
$3 = \dfrac{2x}{72y}$

ganeshie8 · Answer

ganeshie8 · Answer

The solutions you've got for $\lambda$ are correct.

ganeshie8 · Answer

You may use them to find the $x,y$ values of the critical points.

dillonmerv · Answer

So if I equate 3x+y because now I can find the value of x and y through λ, I get ± .0092450033

dillonmerv · Answer

what is this number? what do I do with this?

ganeshie8 · Answer

You get two points : 

x = 0.9985
y = 0.0092

and

x = -0.9985
y =  -0.0092

ganeshie8 · Answer

Those two points lie on the given ellipse x^2+36y^2=1  and maximize/minimize the given function f(x,y) = 3x+y.

ganeshie8 · Answer

Max/min value of 3x+y can only occur at those two points on the ellipse. It cannot occur anywhere else.

dillonmerv · Answer

Got it Got it!!

dillonmerv · Answer

I should've known that.
Thanks

dillonmerv · Answer

So I'll just plug those points into 3x+y and I'll get max and min.

dillonmerv · Answer

The negative should be min