Answer:
Explanation:
Given the function,
F(x, y) = (10x−x²)(4y−y²)
To find the critical point, we will set F x = 0 and Fy= 0
F x
Let U=(10x - x²) and V = (4y - y²)
∂U/∂x = 10-2x
∂V/∂x = 0
∂F/∂x = U•∂V/∂x+ V∂U/∂x
∂F/∂x = (10-2x)•0 + (4y-y²)•(10-2x)
∂F/∂x = (10-2x)(4y-y²)
So, F x=0
Then, (10-2x)(4y-y²)=0
For the multiplication of two number be zero, it is either one of the number is zero or both number is zero.
Then, (10-2x)(4y-y²)=0
(10-2x)=0 OR (4y-y²)=0
10=2x. OR. y(4-y) =0
x=10/2. OR. y=0 or 4-y = 0
x=5. OR. y =0 or y = 4
Fy:
Let U=(10x - x²) and V = (4y - y²)
∂U/∂y = 0
∂V/∂y = 4-2y
∂F/∂y = V•∂U/∂y + U•∂V/∂y
∂F/∂y = (4y-y²)•0 + (10x-x²)•(4-2y)
∂F/∂y = (10x-x²)•(4-2y)
Then, Fy=0
(10x-x²)•(4-2y)=0
So,
(10x-x²)= 0. --------- OR. (4-2y)=0
x(10-x) = 0. --------- OR 4=2y
x=0. Or (10-x) = 0.-- OR y = 4/2
x= 0 or x =10. ---------- OR. y =2
The critical points are (5,2) (0,4) (10,0)
The saddle point D
D = f xx•fyy- (f xy)²
F x = (10-2x)(4y-y²)
Then, F xx = -2(4y-y²)
Also, F xy = (10-2x)(4-2y)
Fy= (10x-x²)•(4-2y)
Then, Fyy = -2(10x-x²)
So, D = f xx•fyy- (f xy)²
D=-2(4y-y²)•(-2(10x-x²)-(10-2x)²(4-2y)²
D = 4xy(4-y)(10-x) -(10-2x)²(4-2y)²
At the point
(5,2)
D=4xy(4-y)(10-x) -(10-2x)²(4-2y)²
D=4•5•2(4-2)(10-5)-(10-2×5)²(4-2×2)²
D =40•-2•5
D=-400
D<0 then, (5,2) is a saddle point
(0,4)
D=4xy(4-y)(10-x) -(10-2x)²(4-2y)²
D=4•0•4(4-4)(10-0)-(10-2×0)²(4-2×4)²
D = 0 - 8²(-4)²
D= -1024
D<0, this point is a critical point
(10,0)
D=4xy(4-y)(10-x) -(10-2x)²(4-2y)²
D=4•10•0(4-0)(10-10)-(10-2×10)²(4-2×0)²
D= - (-10)²(8)²
D = -6400
D<0, this is also a saddle point.