# Function's maxima and minimaConsider f(x; y) = (e^y - e)(x - 2) on the set A = {(x; y) : 0 <= x <= 3; 0<= y <=ln(x+1)} Establish whether f has on A maximum and minimum, and if yes,...

Function's maxima and minima

Consider f(x; y) = (e^y - e)(x - 2) on the set A = {(x; y) : 0 <= x <= 3; 0<= y <=ln(x+1)}

Establish whether f has on A maximum and minimum, and if yes, compute them.

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You need to find the partial derivatives `f_x` and `f_y` using the product rule such that:

`f_x = (e^y - e)'(x - 2) + (e^y - e)(x - 2)'`

`f_x = 0 + e^y - e => f_x = e^y - e`

`f_y = (e^y - e)'(x - 2) + (e^y - e)(x - 2)'`

`f_y = (e^y)(x - 2) + 0 => f_y = (e^y)(x - 2)`

You need to remember that the critical points need to satisfy the system of equations `f_x = 0` and `f_y = 0` such that:

`{(e^y - e = 0),((e^y)(x - 2) = 0):} => {(e^y = e),(x - 2)= 0):}`

`{(y = 1),(x = 2):}`

You need to evaluate the second order partial derivatives such that:

`f_(x x) = (f_x)_x = 0`

`f_(y y) = (f_y)_y = (e^y)(x - 2)`

`f_(x y) = e^y`

You need to evaluate `D = f_(x x)(2,1)f_(y y)(2,1) - f^2_(x y)(2,1)` such that:

`D = 0*(e^1)(2-1) - e^2 = -e^2`

**Since `D= -e^2 <0` , the function has a saddle point at (2,1), over the interval `[0,3].` **