Determine the local extrema for this function,
f(x,y) = √(56x^2-8y^2-16x-31) + 1 - 8x,
whether a local minimum, local maximum or saddle point.
f(x,y) = √(56x^2-8y^2-16x-31) + 1 - 8x,
To determine the critical points, we need the 1st derivatives.
fx = (112x – 16)/{2√(56x^2-8y^2-16x-31)) – 8
fy = -8y/(√(56x^2-8y^2-16x-31))
Setting fx = 0 and fy = 0,
(112x – 16)/{2√(56x^2-8y^2-16x-31)) – 8 = 0 ----------------------- (1)
-8y/(√(56x^2-8y^2-16x-31)) = 0 ----------------------------------------- (2)
Eqn (2) is satisfied for y = 0.
Setting y = 0 in (1),
(112x – 16)/{2√(56x^2-16x-31)) = 8
(112x – 16) = 16√(56x^2-16x-31)
Squaring both sides,
12544x^2 – 3584x + 256 = 256(56x^2-16x-31)
12544x^2 – 3584x + 256 = 14,336x^2 – 4096x – 7936
0 = 1792x^2 – 512x – 8192
The solutions to this quadratic equation are,
x = 16/7, x = -2
Upon checking these solutions, we find that x = -2 is a solution to the quadratic eqn, but does not make f(-2, 0) = 0, as required, and as does f(16/7, 0) =0.
Therefore ignore the solution, x = -2.
We have one critical point ant that is at: (x,y) = (16/7, 0)
For the Hessian, we need the 2nd derivatives.
fxx = (-1/4)(112x – 16)^2/{(56x^2-8y^2-16x-31)^(3/2)) + 56/(√(56x^2-8y^2-16x-31))
fxy = fyx = 4y(112x – 16)/( (56x^2-8y^2-16x-31)^(3/2))
fyy = -64y^2/( (56x^2-8y^2-16x-31)^(3/2)) – 8/(√(56x^2-8y^2-16x-31))
At the critical point (16/7, 0), substituting for x = 16/7 and y = 0 in the expressions for fxx, fxy, fyx, fyy, we get for the Hessian
H = | -8/15 0 |, and det(H) = (-8/15)*(-8/15) = 64/225
| 0 -8/15 |
Since fxx < 0 and det(H) > 0, then the point is a local maximum