# finding maximum and minimum points of a two variable function

• Nov 17th 2012, 02:05 PM
asilvester635
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?
• Nov 17th 2012, 04:06 PM
Dark Sun
Re: finding maximum and minimum points of a two variable function
What was the question asking for exactly?
• Nov 17th 2012, 04:23 PM
asilvester635
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
Attachment 25764
• Nov 17th 2012, 05:36 PM
Prove It
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.
• Nov 17th 2012, 06:37 PM
asilvester635
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?
• Nov 17th 2012, 06:45 PM
asilvester635
Re: finding maximum and minimum points of a two variable function
PLEASE ANYONE???!!!!!! im very confused.. :(((
• Nov 17th 2012, 06:46 PM
hollywood
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
• Nov 17th 2012, 07:03 PM
asilvester635
Re: finding maximum and minimum points of a two variable function
so minimum is (1,1) and maximum is (1,3)????
• Nov 17th 2012, 10:06 PM
asilvester635
Re: finding maximum and minimum points of a two variable function
oops sorry i didn't read the whole thing lol...
• Nov 17th 2012, 10:24 PM
hollywood
Re: finding maximum and minimum points of a two variable function
No problem.

- Hollywood