Consider the function f(x, y)=(10x−x2)(4y−y2). Find and classify all critical points of the function. If there are more blanks than critical points, leave the remaining entries blank.

Step-by-step 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.

b.an increase in a causes b to decrease