So,

So,

and

So,

So,

Given function :

Now, we can check value of the function Z at (1,1), (1,5) , (4,1), (4,5)

So, Z is maximum at (4,1) and maximum value is 29/17. Similarly, Z is minimum at (1,5) and minimum value is 12/9.

The function Z has No local maxima or local minima in the interval ; except at the corner points of the interval.

To see this we can apply the method "Max/min for functions of two variables" which is a tedius task.

First we find partial derivatives w.r.t x and y.

∂f /∂x =

∂f /∂y =

A point (a, b) which is a maximum, minimum or saddle point is called a stationary point.

To find the stationary points of Z = f(x, y), work out ∂f /∂x and ∂f /∂y and set both to zero. This gives you two equations for two unknowns x and y. Solve these equations for x and y

So, to find stationary points, we put both equal to zero and find stationary points.

means x = 0 ; means y = 0

So, only stationary point possible is But at the function Z is Not defined, so, we can say that there is No stationary point for this function.

http://personal.ee.surrey.ac.uk/Personal/S.Zelik/teach/calculus/max_min_2var.pdf