How to find the maximum value of $f(x,y) = 39 - x^2 - y^2$ on the line $x+6y=37$?

so I'm totally lost how to find this question

find the maximum value of $f(x,y) = 39 - x^2 - y^2$ on the line $x+6y=37$

I know how to find local max and min of multivariable equations, but now I don't know what to do given that line equation.

If someone could help me it would be very much appreciated

Thanks in advance

2 Answers

Lagrange multipliers are a nice way to solve any problem of the form: Find the [maximum/minimum/extrema] of [a function] along [a curve].

In this case, we are trying to find the maximum of the function $f(x,y)=39-x^2-y^2$ over the curve $x+6y=37$.

The first step to solve anything using Lagrange multipliers is to find a new function $g(x,y)$ such that our curve, $x+6y=37$, is the level curve $g(x,y)=0$. The method works for any such function, but an easy way to find a good function is to simply move all the terms to the left and make that our $g$:

$ x+6y=37 \\x+6y-37=0 \\g(x,y)=x+6y-37 $

It is clear that this function satisfies our condition because if we set $g(x,y)=x+6y-37=0$, we can rearrange to get $x+6y=37$.

Next comes the "multiplier" part of Lagrange multipliers. Lagrange tells us that any extrema of $f$ along our curve must have the property $\triangledown f(x,y)=\lambda\triangledown g(x,y)$, so every point satisfying that condition along our curve may be (but is not guaranteed to be) an extrema of $f$.

To find these points, we must find the gradient vectors $\triangledown f(x,y)$ and $\triangledown g(x,y)$:

$ f_x(x,y)=\frac{\partial}{\partial x}\left[29-x^2-y^2\right]=-2x \\ f_y(x,y)=\frac{\partial}{\partial y}\left[29-x^2-y^2\right]=-2y \\ \triangledown f(x,y)=(f_x(x,y),f_y(x,y))=(-2x,-2y) $

$ g_x(x,y)=\frac{\partial}{\partial x}\left[x+6y-37\right]=1 \\ g_y(x,y)=\frac{\partial}{\partial y}\left[x+6y-37\right]=6 \\ \triangledown g(x,y)=(g_x(x,y),g_y(x,y))=(1,6) $

Now we can simply solve for points satisfying $\triangledown f(x,y)=\lambda\triangledown g(x,y)$ which are along our curve:

$ \begin{cases} -2x=\lambda \\ -2y=6\lambda \\ x+6y=37 \end{cases} $

This gives us $\lambda=-2$, $x=1$, and $y=6$. The value of $\lambda$ doesn't really matter for our purposes, but this tells us that $(x,y)=(1,6)$ is a potential extrema of $f$ along this curve. If our domain curve weren't infinitely long, the ends of our curve would also be potential extrema, but that's not an issue here.

Since we only generated one point from our Lagrange multiplier, either $f(1,6)$ is the maximum of $f$ along this curve and there is no minimum or vice versa. We can easily check by finding that $f(1,6)=2$ and checking the value of $f$ for some other point along the curve. For example, $f(7,5)=-35<f(1,6)$ so $f(1,6)=2$ must be the maximum of $f$ along the curve $x+6y=37$.

P.S. Sorry for the late reply, hopefully this will help someone in the future.

HINT

One obvious approach: on that line, $x=37-6y$, which you can plug into $f(x,y)$, reducing it to $f(x(y),y)$, a function of one variable, and use ordinary Calculus.

Second approach would be to use something like Lagrange multipliers.