You ought to get this one!!

( This solution is performed using the "Double Differentiation Test for Maxima and Minima". There are Several other ways to solve this problem as well ) $f(x,y)= x^(3) + y^(3) - 12x -3y + 16$

Now, first we partially differentiate f(x,y) with respect to x.

$\frac {df(x,y)}{dx} = \frac {d (x^{3} + y^{3} - 12x -3y + 16)}{dx} \Rightarrow f_{x} = 3x^{2} -12$ .......[1]

Now, we partially differentiate f(x,y) with respect to y.

$\frac {df(x,y)}{dy} = \frac {d (x^{3} + y^{3} - 12x -3y + 16)}{dy} \Rightarrow f_{y} = 3y^{2} - 3$ .......[2]

Now, we equate both equations [1] & [2] to zero and get these results: $x=\pm 2$ ...........[3] $y=\pm 1$ ...........[4]

Now, we obtain the $2^{nd}$ partial derivative of f(x,y) seperately with respect to x & y. So, $f_{xx} = 6x$ .........[5] $f_{yy} = 6y$ ..........[6]

Substituting the values of x and y from equations [3] & [4] in equations [5] & [6] respectively. We get,

$f_{xx}(2) = 12 >0 \rightarrow Point of Minima$ $f_{xx}(-2) = -12 <0 \rightarrow Point of Maxima$

Similarly,

$f_{yy}(1) = 6 >0 \rightarrow Point of Minima$ $f_{yy}(-1) = -6 <0 \rightarrow Point of Maxima$

Thus, the point of local maxima for f(x,y) is x=-2, y=-1 in agreement with the Constraint $( -3\leq x,y \leq3 )$ . Substing the above value of point of Maxima in f(x,y) we'll get the maximum value of the function.

$\Rightarrow f(-2,-1) = (-2)^(3) + (-1)^(3) - 12\times (-2) -3\times (-1) + 16$ $\Rightarrow f(-2,-1) = -8 -1 + 24 + 3 + 16$ $\Rightarrow$ f(-2,-1) = 34 $\rightarrow Local Maxima$

Now, lets check for the boundary points as well. $f(-3,-3) = 7 <34$ $f(3,3) = 25 <34$ $f(-3,3) = 45 >34$ $f(3,-3) = -11 <34$

Now Lets check by taking one boundary point and one point that we calculated out in equations [3] & [4]. $f(3,1) = 5 <34$ $f(3,-1) = 10 <34$ $f(2,3) = 17 <34$ $f(2,-3) = -18 <34$ $f(-2,3) = 50 >34$ $\Leftarrow \Leftarrow$ $f(-2,-3) = 14 <34$

Therefore, point (-2,3) is the Point of Global Maxima and the Globally Maximum value of this function is 50 .