Find the Local Maxima and Minima f(x)=-2x^2-4y^2+2x+2y+3

Problem

ƒ(x,y)=−2*x2−4*y2+2*x+2*y+3

Solution

1. Find the first partial derivatives of the function with respect to x and y

∂(ƒ)/∂(x)=−4*x+2

∂(ƒ)/∂(y)=−8*y+2

2. Set the partial derivatives to zero to find the critical points.

−4*x+2=0⇒x=1/2

−8*y+2=0⇒y=1/4

The critical point is (1/2,1/4)

3. Find the second partial derivatives to use the Second Derivative Test.

(ƒ_x*x)=∂2(ƒ)/(∂(x)2)=−4

(ƒ_y*y)=∂2(ƒ)/(∂(y)2)=−8

(ƒ_x*y)=∂2(ƒ)/(∂(x)*∂(y))=0

4. Calculate the discriminant D using the formula D=(ƒ_x*x)*(ƒ_y*y)−((ƒ_x*y))2

D=(−4)*(−8)−(0)2=32

5. Apply the Second Derivative Test at the critical point.
   Since D>0 and (ƒ_x*x)<0 the function has a local maximum at (1/2,1/4)

6. Evaluate the function at the critical point to find the maximum value.

ƒ(1/2,1/4)=−2*(1/2)2−4*(1/4)2+2*(1/2)+2*(1/4)+3

ƒ(1/2,1/4)=−1/2−1/4+1+1/2+3=3.75

Final Answer

Local Maximum: *ƒ(1/2,1/4)=15/4

Want more problems? Check here!