calculate max and min stationary point and values

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 can switch to polar coordinates and the constraint becomes r = 4, which is easy to substitute. Recall that x = r cos θ and y =r sin θ or x = 4 cos θ and y =4 sin θ

Thus, 6xy - 24(x+y) becomes

F(θ) = 96 sinθ cosθ - 96 cos θ - 96 sin θ = 96(sinθ cosθ - cos θ - sin θ )

dF/dθ = 96 cos²θ - 96 sin²θ + 96 sin θ - 96 cos θ = 0

96[(cosθ - sinθ)(cosθ+sinθ) - (cosθ - sinθ)] = 0

96(cosθ - sinθ)(cosθ+sinθ - 1) = 0

We have two solutions from cosθ - sinθ = 0, π/4 and 5π/4

cosθ+sinθ - 1 = 0 at 0 and π/2

At 0, (x, y) = (4, 0) and F(x, y) = 96(0 - 1 - 0) = -96

At π/4, (x, y) = (2√2, 2√2) and F(x, y) = 96(1/2 - √2) = 48 - 96√2

At π/2, (x, y) = (0, 4) and F(x, y) = 96(0 - 0 - 1) = -96

At 5π/4, (x, y) = (-2√2, -2√2) and F(x, y) = 96(1/2 + √2) = 48 + 96√2

a) Maximum point: (-2√2, -2√2)

Maximum value: 48 + 96√2

b) Two minimum points

(4, 0)

F(x,y) = -96

(0, 4)

F(x,y) = -96

c)

(2√2, 2√2)

F(x, y) = 48 - 96√2