stationary points for max and min

Question

for F(x,y) = 6x . y - 24(x + y)

Subject to x^2 + y^2 = 16

Find:

a) a global maximum stationary point and its value

Maximum point: (x,y)max =

Maximum value: F(x,y)max =

b) a global minimum stationary point and its value:

Minimum point (x,y)min =

Minimum value: F(x,y)min =

c) another stationary point that is not a global maximum or minimum and it’s value

(x,y) =

F(x,y) =

Answer 1

We use Lagrange multiple methods to approach this question. Let g(x,y)=x^2+y^2-16. Then

grad(F)=r*grad(g) gives <6y-24, 6x-24> = r*<2x, 2y>, that is,

6y-24 = 2rx (1)

6x-24 = 2ry (2)

x^2+y^2 = 16 (3)

(1)-(2) yields 6(y-x) = 2r(x-y) which gives 2(r+3)(x-y) =0, that is, x=y or r=-3.

When x=y, (3) gives x^2+x^2=16 or x^2=8, so x=-2sqrt(2) or 2sqrt(2). So we have (2sqrt(2), 2sqrt(2)) and (-2sqrt(2), -2sqrt(2)).

When r=-3, (1) yields y=4-x. Plugging it into (3) yields x^2 + (4-x)^2 =16 or 2x(x-4)=0, that is x=0 or x=4.

So y=4 or y=0. So we have (0, 4) and (4,0).

Now we have 4 different stationary points (2sqrt(2), 2sqrt(2)), (-2sqrt(2), -2sqrt(2)), (0, 4), and (4,0).

Therefore, F(2sqrt(2), 2sqrt(2))=6(2sqrt(2))(2sqrt(2))-24(2sqrt(2)+2sqrt(2))=48-96sqrt(2),

F(-2sqrt(2), -2sqrt(2)) = 6(-2sqrt(2))(-2sqrt(2))-24(-2sqrt(2)-2sqrt(2))=48+96sqrt(2),

F(4,0) = 6(4)(0) - 24(4+0) = -96, and

F(0,4) = 6(0)(4) - 24(0+4) = -96.

(a) The global maximum stationary point is (-2sqrt(2), -2sqrt(2)) and the maximum value is F(-2sqrt(2), -2sqrt(2)) = 48+96sqrt(2).

(b) The global minimum stationary points are (4, 0) and (0, 4) and the minimum value is -96.

(c) The other stationary point that is neither a global maximum nor a global minimum is (2sqrt(2), 2sqrt(2)), and the corresponding value of F(x,y) is 48-96sqrt(2).