Finding stationary points

Question

Need answers to this question. It is one question but in two parts.

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

Subject to x^2 + y^2 = 16

Find:

a) a global 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

I am assuming a typo: question a) should be

"a) a global maximum stationary point and its value"

To find the stationary points we use the method of Lagrange multipliers,

forming the function

G(x,y) = 6xy - 24(x+y) - λ(x² + y²)

and sets its two partial derivatives to 0

6y - 24 - 2λx = 0

6x - 24 - 2λy = 0

From the first equation express λ:

λ = (6y - 24)/2x

and plug it into the second equation

6x - 24 - (6y - 24)y/x = 0

Simplifying the above:

6x² - 24x - 6y² + 24y = 0

(x² - y²) - 4(x - y) = 0

(x + y)(x - y) - 4(x - y) = 0

(x + y -4)(x - y) = 0

The above has two types of solutions:

x - y = 0, (1)

x + y - 4 = 0 (2)

When we combine (1) with the constraint x² + y² = 16

we get two solutions

x = y = √8

x = y = -√8

When we combine (2) with the constraint x² + y² = 16

we get two solutions

x = 0, y = 4

x = 4, y = 0

So in total we get four stationary points with the following F-values

F(√8, √8) = 48(1 - √8) = -87.76

F(-√8, -√8) = 48(1 + √8) = 183.76

F(0, 4) = -96

F(4, 0) = -96

The answers are

a) F(-√8, -√8) = 183.76

b) F(0, 4) = -96 and F(4, 0) = -96

c) F(√8, √8) = -87.76