# I am so lost with this question please help Question : f(x,y)=4x^(3) + y^3 - 6x^2 -6y^2 + 5 find all critical points of the given function f(x,y) and use the second partials test to classify as a...

I am so lost with this question please help

Question : f(x,y)=4x^(3) + y^3 - 6x^2 -6y^2 + 5

find all critical points of the given function f(x,y) and use the second partials test to classify as a relative maximum, a relative minimum or a saddle

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You need to remember that you may find the critical points of a two variable function solving the system of equations `f_x(x,y) = 0` and `f_y(x,y) = 0` .

You need to find the first order partial derivatives such that:

`f_x(x,y) = (del(4x^3 + y^3 - 6x^2 -6y^2 + 5 ))/(del x)`

Notice that you need to differentiate with respect to x considering y as a constant such that:

`f_x(x,y) = 12x^2 - 12x`

`f_y(x,y) = (del(4x^3 + y^3 - 6x^2 -6y^2 + 5 ))/(del y)`

`f_y(x,y) = 3y^2 - 12y`

You need to solve the system of equations such that:

`{(f_x(x y)=0),(f_y(x y)=0):}` => `{(12x^2 - 12x = 0),(3y^2 - 12y = 0):}` =>`{(x^2-x = 0),(y^2 - 4y = 0):}` `=gt{(x(x-1)=0),(y(y-4)=0):}` =>`{(x=0,x=1),(y=0,y=4):}`

**Hence, evaluating the critical points of the function yields (0,0), (0,4) , (1,0) , (1,4).**

You need to classify the critical points, hence, you need to evaluate D such that:

`D = f_(x x)(a,b)f_(y y)(a,b) - f^2_(x y)(a,b)`

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

`f_(x x) (x,y) = (del(12x^2 - 12x))/(del x)`

`f_(x x) (x,y) = 24x - 12`

`f_(x y) (x,y) = (del(12x^2 - 12x))/(del y)`

`f_(x y) (x,y) = 0`

`f_(x y ) (x,y) = (del(3y^2 - 12y))/(del y)`

`f_(x y ) (x,y) = 6y - 12`

You need to evaluate D for the first critical point (0,0) such that:

`D_(0,0) = (24*0 - 12)(6*0 - 12) - 0`

`D_(0,0) = 144`

**Notice that `D_(0,0) = 144 > 0` and `f_(x x) (0,0) = -12 < 0` , hence, the function has a relative maximum at (0,0).**

You need to evaluate D for the first critical point (0,4) such that:

`D_(0,4) = (24*0 - 12)(6*4 - 12) - 0`

`D_(0,4) = -144`

**Since `D_(0,4) = -144 < 0` , hence, the function has a saddle point at (0,4). **

You need to evaluate D for the first critical point (1,0) such that:

`D_(1,0) = (24*1 - 12)(6*0 - 12) - 0`

`D_(1,0) = -144`

**Since `D_(1,0) = -144 < 0` , hence, the function has a saddle point at (1,0).**

You need to evaluate D for the first critical point (1,4) such that:

`D_(1,4) = (24*1 - 12)(6*4 - 12) - 0`

`D_(1,4) = 144`

**Since `D_(1,4) = 144 > 0` and `f_(x x) (0,0) = 12> 0` , hence, the function has a relative minimum at (1,4).**