Suppose f(x,y) = A - (x^2 + Bx + y^2 +Cy). What values of A, B, and C give f(x,y) a local maximum value of 15 at the point (-2,1)?
Any help would be appreciated!
To find the values of A, B, and C that give the function f(x, y) a local maximum value of 15 at the given point (-2, 1), you need to use the concept of critical points and the second partial derivative test.
First, find the partial derivatives of f(x, y) with respect to x and y:
∂f/∂x = -2x - B
∂f/∂y = -2y - C
Next, find the critical point by setting these partial derivatives equal to zero:
-2x - B = 0 ⇒ -2x = B (equation 1)
-2y - C = 0 ⇒ -2y = C (equation 2)
Now, substitute the x and y values of the given point (-2, 1) into equations 1 and 2:
-2(-2) = B ⇒ B = 4
-2(1) = C ⇒ C = -2
So, the values of B and C are 4 and -2, respectively.
To determine the value of A, substitute the x, y, and B, and C values into the function f(x, y) and set it equal to the local maximum value of 15:
f(x, y) = A - (x^2 + Bx + y^2 + Cy)
15 = A - ((-2)^2 + 4(-2) + 1^2 + (-2)(1))
15 = A - (4 - 8 + 1 - 2)
15 = A - (-3)
Finally, solve for A:
15 = A + 3
A = 12
Therefore, the values of A, B, and C that give f(x, y) a local maximum value of 15 at the point (-2, 1) are A = 12, B = 4, and C = -2.