# Given:C(x,y)=2x^2+3y^2+2xy-6x-18y+54 Find a) Cx(x,y) and Cy(x,y)  b)xand y such that Cx(x,y)=0 and Cy(x,y)=0  x and y concented to the C's are small

mlehuzzah | Certified Educator

To find `C_x` , treat x as a variable and y as a constant, and find the derivative:

`2x^2+3y^2+2xy-6x-18y+54`

x is the variable, then the derivative of `2x^2` is `4x`

if y is a constant, then `y^2` is a constant, and `3y^2 is a constant, so the derivative of `3y^2` is 0

`2xy` is really the variable `x` times the constant `2y`; the derivative of a constant times the variable is the constant, so the derivative is `2y`

similarly, the derivative of `-6x` is -6, the derivative of `-18y` is 0, and the derivative of 54 is 0.

So:

`C_x(x,y) = 4x+0+2y-6-0+0=4x+2y-6`

To find `C_y`, treat y as the variable and x as a constant, and find the derivative:

`2x^2+3y^2+2xy-6x-18y+54`

Now, x is a constant, so `2x^2` is a constant, and the derivative is 0

The derivative of `3y^2` is `6y`

`2xy` is the variable `y` times the constant `2x` so the derivative is `2x`

x is a constant, so `-6x` is a constant, and its derivative is 0

the derivative of `-18y` is -18, and the derivative of -6 is 0

So:

`C_y(x,y)=0+6y+2x-0-18+0=6y+2x-18`

Now, we want to find x and y such that both `C_x` and `C_y` are 0:

So:

`0=4x+2y-6`

`0=2x+6y-18`

This is a system of equations. We can solve it by multiplying the bottom equation by 2:

`0=4x+2y-6`

`0=4x+12y-36`

Subtracting, we get:

`0=-10y+30`

so `y=3`

plugging this in (to either equation) we get:

`0=4x+2(3)-6`

`0=4x`

`x=0`

So if `x=0` and `y=3`, then `C_x(x,y)=C_y(x,y)=0`