Mdhiggenz said:Homework Statement
Locate all relative max,min,and saddle points if any.
f(x,y)=x2+xy+y2-3x
fx=2x+y-3
fy=x+2y
Skipping some algebra I get the critical points (2,-1)
fxx=2
fyy=2
d=fxx*fyy-f(x0,y0)2
d=4-9=-5
I know I'm messing up at f(x0,y0)
I'm simply plugging in my c.p points (2,-1) into
f(x,y)=x2+xy+y2-3x
but seem to keep getting -3
While the value 1 is suppose to come out
Homework Equations
The Attempt at a Solution
Mdhiggenz said:I'm a bit confused what do you mean mixed second derivative?
A relative maximum is a point on a graph where the function has the highest value within a certain interval. This means that there are points both to the left and right of the maximum where the function has a lower value.
To find the relative maximum of a function, you must first take the derivative of the function and set it equal to 0. Then, solve for the variable to find the critical points. Next, plug the critical points into the original function to determine the values at those points. The highest value among the critical points is the relative maximum.
Yes, a function can have multiple relative maxima. This occurs when the function has multiple peaks within an interval. Each peak will have its own relative maximum value.
A relative maximum is a point where the function has the highest value within an interval, while an absolute maximum is the highest value of the entire function. A relative maximum may occur at a critical point, but an absolute maximum can only occur at an endpoint of the interval or at a point where the derivative is undefined.
Finding relative maxima is important in Calculus 3 because it allows us to determine important information about the function, such as the behavior and rate of change. It also helps us to identify important points on the graph and can be useful in optimization problems.