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Calculus/Help me with the extreme values of the function

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Question
z = x2 – y + 4ln(3-x) + ln(y+2)

This was problem for the exam.. and I didn't know how to solve it. Can you help me? Thanks in advance.

Answer
z=x²–y+4ln(3-x)+ln(y+2)
1st we need to find the partial derivatives :
∂z/∂x = 2x-[4/(3-x)]
∂z/∂y = [1/(y+2)]-1
We solve the set of equations : ∂z/∂x=0 & ∂z/∂y=0 :
2x-[4/(3-x)]=0 -> x²-3x+2=0 -> { x1=2 & x2=1 }
[1/(y+2)]-1=0 -> { y1=-2 }
So points of extremum are : (2,-2) & (1,-2) .
To know if each is max , min or saddle, we have to calculate :
∂²z/∂x² , ∂²z/∂y² , ∂²z/∂x∂y .

Alon.

Calculus

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Alon Mandes

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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

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