wowmaths said:Homework Statement
f(x,y) = xy(9x^2 + 3y^2 -16)
Find the critical points of the function and their nature (local maximum, local minimum or saddle)Homework Equations
The Attempt at a Solution
I have partially differentiated the equation into:
fx = 27yx^2* + 3y^3 -16y
fy = 9x^3 + 9xy^2 - 16x
How do I go on from there though?
Zondrina said:Hey there, welcome to the forum.
So, recall that a critical point of a function f(x) occurs when the derivative f'(x) = 0.
Now translating that to the multivariate case is not very different. Set fx = 0 and fy = 0 and then solve for your critical points.
Ray Vickson said:You solve the equations
27*y*x^2 + 3*y^3 - 16*y = 0
9*x^3 + 9*x*y^2 - 16*x = 0.
The first one says either y = 0 or 27*x^2 + 3*y^2 - 16 = 0, and this last one is the equation of an ellipse in (x,y) space. The second one says either x = 0 or 9*x^2 + 9*y^2 -16 = 0. What kind of geometric figure does that describe in (x,y)-space?
RGV
wowmaths said:Thanks a lot for the welcome and the help. Greatly appreciated.
Just a question:
For both the equations:
fx = 27*y*x^2 + 3*y^3 -16*y = 0
and
fy = 9*x^3 + 9*x*y^2 - 16*x = 0
Am I supposed to solve both equations simultaneously to find the stationary points?
I tried using a calculator to solve both equations simultaneously and I've gotten 9 stationary points.
Am I on the right track?
A critical point is a point in a multivariable function where the partial derivatives are equal to zero or do not exist. This means that the function is not changing in any particular direction at that point.
To find critical points, you must first take the partial derivatives of the function with respect to each variable. Then, set each partial derivative equal to zero and solve the resulting system of equations to find the values of the variables at the critical point.
Critical points help us identify the location of local extrema (maximum or minimum values) in a multivariable function. They also help us determine the nature of these extrema, whether they are local maxima, local minima, or saddle points.
To determine the nature of a critical point, we use the second derivative test. This involves taking the second partial derivatives of the function and evaluating them at the critical point. If the resulting value is positive, the critical point is a local minimum. If the value is negative, the critical point is a local maximum. If the value is zero, the second derivative test is inconclusive and further analysis is needed.
Yes, a multivariable function can have multiple critical points. In fact, it is common for functions to have multiple critical points. These points may represent local extrema or saddle points.