# Find the absolute extrema of the given function on the indicated closed and bounded set R. f(x,y)...

## Question:

Find the absolute extrema of the given function on the indicated closed and bounded set {eq}R {/eq}.

{eq}f(x,y) = x^{2} - 4y^{2} - 4x + 16y\,; \quad R \, {/eq} is the square region with vertices: {eq}\; (0,0),\; (0,7),\; (7,7),\; (7,0) {/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.

## Answer and Explanation:

We have the function {eq}f(x,y) = x^{2} - 4y^{2} - 4x + 16y\,; \quad R \, {/eq} is the square region with vertices: {eq}\; (0,0),\; (0,7),\; (7,7),\; (7,0) {/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=2x-4 \ and \ f_y=-8y+16-----(i) {/eq}

We have given the square region with vertices: {eq}\; (0,0),\; (0,7),\; (7,7),\; (7,0) {/eq}

This implies {eq}0 \leq x \leq 7 \ and \ 0 \leq y \leq 7 {/eq}

On equating the partial derivatives with 0, we get,

{eq}x=2 \ and \ y=2 {/eq}

Thus, (2,2) is the critical point.

And at the boundary points

When x=0, we get,

{eq}f(x,y) = x^{2} - 4y^{2} - 4x + 16y \\ f(0,y) = - 4y^{2} + 16y {/eq}

This gives y=2 is the critical point.

**Thus, (0,2) is the point of evaluation.**

When x=7, we get,

{eq}f(x,y) = x^{2} - 4y^{2} - 4x + 16y \\ f(7,y) = 49- 4y^{2} -28 + 16y {/eq}

This gives y=2 is the critical point.

**Thus, (7,2) is the point of evaluation.**

When y=0, we get,

{eq}f(x,y) = x^{2} - 4y^{2} - 4x + 16y \\ f(x,0) = x^{2} - 4x {/eq}

This gives x=2 is the critical point.

**Thus, (2,0) is the point of evaluation.**

When y=7, we get,

{eq}f(x,y) = x^{2} - 4y^{2} - 4x + 16y \\ f(x,0) = x^{2} -196 - 4x +112 {/eq}

This gives x=2 is the critical point.

**Thus, (2,7) is the point of evaluation.**

Now the values of the function at each point are

{eq}f(2,2) = 2^{2} - 42^{2} - 4(2) + 16(2)=4-16-8+32=12 \\ f(2,0) = 2^{2} - 4(2) =4-8=-4 \\ f(2,7) = 2^{2} - 4(7)^{2} - 4(2) + 16(7)=4-196-8+112=-88 \\ f(0,2) = - 42^{2} + 16(2)=-16+32=16 \\ f(7,2) = 7^{2} - 42^{2} - 4(7) + 16(2)=49-16-28+32=37 {/eq}

**Therefore, the absolute maxima of the given function is at the point (7,2) and the absolute minima of the function is at the point (2,7).**

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from Math 104: Calculus

Chapter 9 / Lesson 3