Jump to content

Minimum and Maximum help


Math_Struggles
 Share

Recommended Posts

First off - some maths teachers would immediately dock points for calling the expression in your first post an equation; without an equals it isn't an equation. Secondly - do you have in your head what sort of equation (assuming it equals zero) this is - ie is it possible to draw a simple line representing this curve on a piece of paper. Thirdly perhaps give a bit more detail about what you did to simplify.

Link to comment
Share on other sites

I've simplified it to the equations

 

0 = x(x-y)^2 and 0 = y(x-y)^2. I know that x and y can't be zero because at (0,0) the function is undefined. So are the critical points anywhere where x = y? That seems to be the place that the other quantity in the two equations would be zero.

If those are indeed the derivatives, that makes sense. I presume you took the gradient and set its components to zero?

Link to comment
Share on other sites

Yes that is what I did. I took the partial derivatives with respect to x and y and thensimplified them down to that result.

 

I'll attemt to show my work here, but I don't know how to make the work look like real math steps rather than typed out math.

 

f(x) = partial derivative with respect to x f(y) = same with y

 

f(X) = 0 = 2((x^2-2xy+3y^2)(3x+y) - (3x^2+2xy+y^2)(x-y))/(x^2-2xy+3y^2)^2

 

f(x) = 0 = (3x^3-6x^2y+9xy^2+x^2y-2xy^2+3y^3-3x^3-2x^2y-xy^2+3x^2y+2xy^2+y^3)/(x^2-2xy+3y^2)^2

 

f(x) = 0 = (-4x^2y+8xy^2+4y^3)/(x^2-2xy+3y^2)^2

 

f(x) = 0 = (y^3+2xy^2-x^2y)/(x^2-2xy+3y^2)^2

 

The only way that f(x) will be zero is if the numerator is zero, so we set it equal to zero:

 

0 = y^3+2xy^2-x^2y

 

0 = y(y^2+2xy-x^2)

 

I realize now that when I wrote y(x-y)^2 I was wrong in that y^2+2xy-x^2 does not factor to (x-y)^2!

 

f(y) = 0 = 2((x^2-2xy+3y^2)(3x+y) - (3x^2+2xy+y^2)(x-y))/(x^2-2xy+3y^2)^2

The work is very similar for f(y), and I end up with

 

0 = x(x^2-2xy-y^2)

 

So now I don't know where to go from there. I see that the f(x) equation can be rewritten as 0 = y(x^2-2xy-y^2), but I don't know how that really helps matters.

 

 

Thanks!

 

 

Link to comment
Share on other sites

Yes that is what I did. I took the partial derivatives with respect to [math] x [/math] and [math] y [/math] and thensimplified them down to that result.

 

I'll attemt to show my work here, but I don't know how to make the work look like real math steps rather than typed out math.

 

[math] f(x) = [/math] partial derivative with respect to [math] x [/math]. [math] f(y) = [/math] same with [math] y [/math]

 

 

[math] f(x) = 0 = 2 \frac{(x^2-2xy+3y^2)(3x+y) - (3x^2+2xy+y^2)(x-y)}{(x^2-2xy+3y^2)^2} [/math]

 

 

[math] f(x) = 0 = \frac{3x^3-6x^2y+9xy^2+x^2y-2xy^2+3y^3-3x^3-2x^2y-xy^2+3x^2y+2xy^2+y^3}{(x^2-2xy+3y^2)^2} [/math]

 

 

[math] f(x) = 0 = \frac{-4x^2y+8xy^2+4y^3}{(x^2-2xy+3y^2)^2} [/math]

 

 

[math] f(x) = 0 = \frac{y^3+2xy^2-x^2y}{(x^2-2xy+3y^2)^2} [/math]

 

 

The only way that [math] f(x) [/math] will be zero is if the numerator is zero, so we set it equal to zero:

 

[math] 0 = y^3+2xy^2-x^2y [/math]

 

 

[math] 0 = y(y^2+2xy-x^2) [/math]

 

 

I realize now that when I wrote [math] y(x-y)^2 [/math] I was wrong in that [math] y^2+2xy-x^2 [/math] does not factor to [math] (x-y)^2 [/math]!

 

 

[math] f(y) = 0 = 2 \frac{(x^2-2xy+3y^2)(3x+y) - (3x^2+2xy+y^2)(x-y)}{(x^2-2xy+3y^2)^2} [/math]

 

 

The work is very similar for [math] f(y) [/math], and I end up with

 

[math] 0 = x(x^2-2xy-y^2) [/math]

 

 

So now I don't know where to go from there. I see that the [math] f(x) [/math] equation can be rewritten as [math] 0 = y(x^2-2xy-y^2) [/math], but I don't know how that really helps matters.

 

 

Thanks!

 

Hi! Latex is used to format your mathematics for our viewing purposes. You can click on any example to get further details.

 

 

This sounds like a simple D-Test or Second Derivative Test. A thorough and comprehensible explanation can be found in any introductory calculus textbook.

Edited by Xittenn
Link to comment
Share on other sites

The only way that f(x) will be zero is if the numerator is zero, so we set it equal to zero:

0 = y(y^2+2xy-x^2)

 

The work is very similar for f(y), and I end up with

0 = x(x^2-2xy-y^2)

 

Well that is quite fortunate in that both those have the same factor, which must be zero if you eliminate the possibility where x and y are zero. Then solving the quadratic equation gives you the other answers.

Link to comment
Share on other sites

Yeah I used the quadratic equation and got

 

[math]

 

y = (-1+\sqrt{2})x

 

[/math]

 

[math]

 

y = -(1+\sqrt{2})x

 

[/math]

 

Now I don't know how to use that to find the critical points. Is it any x and y that fulfill that? And if it's any, meaning there are infinite critical points, how do I know which is the max or min?

Edited by Math_Struggles
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.