# Thread: finding maximum and minimum points of a two variable function

1. ## finding maximum and minimum points of a two variable function

Here's the function
x^2 + 4y^2 - 2x - 16y + 13 = 0

Now time to take the derivative with respect to x, I got
fx = 2x + 0 - 2 - 0
= 2x - 2
Now time to take the derivative with respect to y, I got
fy = 0 + 8y - 0 - 16
= 8y - 16

This is what I got so far and Idk wether I did it right.... is there anything i did wrong?

2. ## Re: finding maximum and minimum points of a two variable function

What was the question asking for exactly?

3. ## Re: finding maximum and minimum points of a two variable function

The book is asking to find the maximum and minimum points on the graph.... (Finding maximum and minimum points with two variable functions)
Here's how it looks like on the book

4. ## Re: finding maximum and minimum points of a two variable function

The critical points are located where both of the partial derivatives are 0.

Then you need to check their nature using the Hessian.

5. ## Re: finding maximum and minimum points of a two variable function

sorry i don't understand the equation..... although i have the work that I've done so far posted up there.... can you please check if i did it right?

6. ## Re: finding maximum and minimum points of a two variable function

PLEASE ANYONE???!!!!!! im very confused.. ((

7. ## Re: finding maximum and minimum points of a two variable function

ProveIt: I think the OP meant to implicitly define a graph in the x-y plane, not a function of two variables.

asilvester635: You have $\frac{dy}{dx}=\frac{\partial{f}/\partial{x}}{\partial{f}/\partial{y}}$, so the minimum and maximum y values are going to occur when $\frac{\partial{f}}{\partial{x}}=0$. This is when x=1, and plugging in to the equation gives:
$x^2+4y^2-2x-16y+13=0$
$1+4y^2-2-16y+13=0$
$4y^2-16y+12=0$
$y^2-4y+3=0$
$y=1,3$
So (1,1) is a minimum and (1,3) is a maximum.

As it turns out, there is a more geometric way to analyze this function. Complete the square in both x and y:
$x^2+4y^2-2x-16y+13=0$
$x^2-2x+4y^2-16y=-13$
$x^2-2x+1+4y^2-16y+16=-13+1+16=4$
$(x-1)^2+4(y-2)^2=4$
so the graph is an ellipse centered at (1,2), and the minimum and maximum points are (1,1) and (1,3) as before.

- Hollywood

8. ## Re: finding maximum and minimum points of a two variable function

so minimum is (1,1) and maximum is (1,3)????

9. ## Re: finding maximum and minimum points of a two variable function

oops sorry i didn't read the whole thing lol...

No problem.

- Hollywood