Setting the minimum point of a quadratic function

Question

I want to set the point $(150, 400)$ as the minimum of the quadratic function:

$$f(x,y)= ax^2 + bxy + cy^2 +dx + ey + g$$

To do this, I solve this set of equations:

$$ \frac{\partial f}{\partial x} = 2ax + by + d = 0 $$ $$ \frac{\partial f}{\partial y} = bx + 2cy + e = 0 $$ $$ \frac{\partial^2 f}{\partial x^2} = 2a > 0 $$ $$ \frac{\partial^2 f}{\partial y^2} = 2c > 0 $$ $$ \frac{\partial^2 f}{\partial x\partial y} = b > 0 $$

by setting $a = c = 1$ and $b = 3$ to meet the above conditions. I can then solve for $d$, $e$, and $g$ by evaluating these equations at the set minimum $(150, 400)$. These values are $-1500$, $-850$, and $-202500$, respectively. However, when I make a contour plot this function, the minimum is nowhere close to the set point (where the black point shows the global minimum of the plot):

Have I done something incorrectly? I can't figure out why I am not seeing a valley at $(150, 400)$.

Answer 1

Maybe you're looking for this polynomial $$f(x,y)=x^2+y^2-300x-800y+182500$$

To get this do nothing, just expand $$(x-150)^2+(y-400)^2$$

Answer 2

The second equation becomes

$$3*150 + 2*400 + e = 0 $$ $$450 + 800 + e = 0$$ $$e = -1250$$

I think you overlooked the $2$ multiplying $c$.