Find equation of tangent plane to$ f(x,y)=2x^2-4xy+6y^2$ have slope $27$ in positive $x$ direction and slope $5$ in positive y direction.

Question

I only got the partial derivative is $4x-4y$ and $-4x+12y$, don't understand why the tangent plane correct answer is $z=27\left(x-\frac{43}4\right)+5(y-4)+\frac{1241}8$? How to get it?

Answer 1

Guide:

Set $$4x_0-4y_0 = 27$$ $$-4x_0+12y_0 = 5$$

Solve for $(x_0, y_0)$.

Also We have to make sure that the tangent plane touches $z=f(x,y)$,

Evaluate $z_0 = f(x_0,y_0)$.

Then the tangent plane is

$$z = 27(x-x_0)+5(y-y_0)+z_0$$

Answer 2

Recall that the linearization of a function $z = f(x,y)$ is the unique linear map whose graph is the tangent plane. It is given by,

$$z = L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$$

A nice geometric characterization is that it is the unique plane which containes the lines $L_x,L_y$. Here $L_y,L_x$ are the images of $(a,b+t)$ and $(a+s,b)$ under $f$ respectively which pass through $(a,b, f(a,b))$ i.e,

$$L_x: f(a,b) + f_x(a,b)(x-a)$$

$$L_y: f(a,b)+f_y(a,b)(y-b)$$

They give you slopes of these lines, $f_x(a,b) = 27$ and $f_y(a,b) = 5$. You need to use this to find the corresponding $(a,b)$.