# Find the absolute maximum and minimum values of f on the set D. f(x,y) = x^{3} - 3x - y^{3} + 12y...

## Question:

Find the absolute maximum and minimum values of {eq}f {/eq} on the set {eq}D {/eq}.

{eq}\displaystyle f(x,y) = x^{3} - 3x - y^{3} + 12y + 5, \quad\quad D\; {/eq} is quadrilateral whose vertices are {eq}\; (-2,3), \; (2,3), \; (2,2), \; {/eq} and {eq}\; (-2,-2) {/eq}

## Absolute extrema

Let f(x,y) be a function defined in a domain D then the function is said to have the absolute extrema at the critical points or at the boundary points if the value of the function is minimum or maximum.

We have the function {eq}\displaystyle f(x,y) = x^{3} - 3x - y^{3} + 12y + 5, \quad\quad D\; {/eq} is quadrilateral whose vertices are {eq}\; (-2,3), \; (2,3), \; (2,2), \; {/eq} and {eq}\; (-2,-2) {/eq}

To find the absolute extrema of the function, we need to check the value of the function at the critical points and the boundary point of the region.

Now, differentiate the given function with respect to x and y, we get,

{eq}f_x=3x^2-3 \ and \ f_y=-3y^2+12-----(i) {/eq}

We have given the quadrilateral whose vertices are {eq}\; (-2,3), \; (2,3), \; (2,2), \; {/eq} and {eq}\; (-2,-2) {/eq}

This implies {eq}-2 \leq x \leq 2 \ and \ -2 \leq y \leq 3 {/eq}

On equating the partial derivatives with 0, we get,

{eq}x=\pm 1 \ and \ y=\pm2 {/eq}

Thus, (1,2), (-1,2), (1,-2) and (-1,-2) are the critical points.

But (-1,2) and (1,2) are the only points that lie in the given quadrilateral.

And at the boundary points

When x=-2, we get,

{eq}f(x,y) = x^{3} - 3x - y^{3} + 12y + 5 \\ f(-2,y) = -8 +6 - y^{3} + 12y + 5 {/eq}

This gives {eq}y=\pm 2 {/eq} are the critical point.

Thus, (-2,2) and (-2,-2) are the point of evaluation.

When x=2, we get,

{eq}f(x,y) = x^{3} - 3x - y^{3} + 12y + 5 \\ f(2,y) = 8 - 6 - y^{3} + 12y + 5 {/eq}

This gives {eq}y=\pm 2 {/eq} is the critical point.

but (2,-2) doesn't lie in the quadrilateral

Thus, (2,2) is the point of evaluation.

When y=-2, we get,

{eq}f(x,y) =x^{3} - 3x - y^{3} + 12y + 5 \\ f(x,-2) = x^{3} - 3x +8 -24 + 5 {/eq}

This gives {eq}x=\pm 1 {/eq} are the critical point.

but (1,-2) and (-1,-2) doesn't lie in the quadrilateral

Thus, there is no point of evaluation.

When y=3, we get,

{eq}f(x,y) = x^{3} - 3x - y^{3} + 12y + 5 \\ f(x,3) = x^{3} - 3x - 27 + 36 + 5 {/eq}

This gives {eq}x=\pm 1 {/eq} are the critical point.

but (-1,3) doesn't lie in the quadrilateral

Thus, (1,3) is the point of evaluation.

Now the values of the function at each point are

{eq}f(-1,2) = x^{3} - 3x - y^{3} + 12y + 5=-1+3-8+24+5=21 \\ f(1,2) =x^{3} - 3x - y^{3} + 12y + 5=1-3-8+24+5=19 \\ f(-2,2) = x^{3} - 3x - y^{3} + 12y + 5=-8+6-8+24+5=19 \\ f(-2,-2) = x^{3} - 3x - y^{3} + 12y + 5=-8+6+8-24+5=-13 \\ f(2,2) = x^{3} - 3x - y^{3} + 12y + 5=8-6-8+24+5=23 \\ f(1,3) = x^{3} - 3x - y^{3} + 12y + 5=1-3-27+36+5=12 {/eq}

Therefore, the absolute maxima of the given function is at the point (2,2) and the absolute minima of the function is at the point (-2,-2). 