# Find the linearization L(x,y) of the function f(x,y) = sqrt(54-4x^2-1y^2) at (3,3).

## Question:

Find the linearization {eq}L(x,y) {/eq} of the function {eq}f(x,y) = \sqrt{54-4x^2-1y^2} \ {/eq} at {eq}\ (3,3) {/eq}.

## Linearization of Function:

Consider the function f(x,y) the linearization of function f(x,y) at point (a,b) is given by the formula {eq}L = f(a,b) + f_x(a,b)(x-a) + f_y(a,b) (y-a)\\ {/eq}. Use this formula to get the linearization of given function. Where {eq}f_x,f_y {/eq} are partial derivatives of function f(x,y) with respect to x and y.

{eq}\displaystyle f(x,y) = \sqrt{54-4x^2-y^2}\\ \displaystyle f(3,3) = \sqrt{54-4x^2-y^2} = \sqrt{54-36-9} = \sqrt{9} = 3\\ {/eq}

Differentiating f(x,y) with respect to x, y

{eq}\displaystyle f_x(x,y) = \frac{d(-4x^2)}{2\sqrt{54-4x^2-y^2}} = \frac{-4x}{\sqrt{54-4x^2-y^2}}\\ \displaystyle f_y(x,y)= \frac{d(-y^2)}{2\sqrt{54-4x^2-y^2}} = \frac{-y}{\sqrt{54-4x^2-y^2}}\\ f_x(3,3) = -4 \\ f_y(3,3) = -1\\ {/eq}

Linearization of f(x,y) at (3,3) is given by

{eq}L = f(3,3) + f_x(3,3)(x-3) + f_y(3,3) (y-3)\\ L = 3 + (-4) (x-3) + (-1) (y-3)\\ L = 18-4x-y {/eq}