To find the maxima, minima, and saddle points of the function f(x,y)=x2+xy+y2+3x+2y+5, we follow these steps:
Step 1: Find the first partial derivatives and set them to zero to find critical points.
The given function is f(x,y)=x2+xy+y2+3x+2y+5.
The partial derivative with respect to x is:
fx=∂x∂(x2+xy+y2+3x+2y+5)=2x+y+3
The partial derivative with respect to y is:
fy=∂y∂(x2+xy+y2+3x+2y+5)=x+2y+2
(Note: The fy in your work was x+2, which is incorrect for the given function. We will use the correct fy=x+2y+2.)
Set fx=0 and fy=0:
- 2x+y+3=0
- x+2y+2=0
From equation (1), we can express y in terms of x:
y=−2x−3
Substitute this into equation (2):
x+2(−2x−3)+2=0
x−4x−6+2=0
−3x−4=0
−3x=4
x=−34
Now substitute x=−34 back into the expression for y:
y=−2(−34)−3
y=38−39
y=−31
The only critical point is (−34,−31).
Step 2: Find the second partial derivatives.
fxx=∂x∂(2x+y+3)=2
fyy=∂y∂(x+2y+2)=2
fxy=∂y∂(2x+y+3)=1
Step 3: Calculate the discriminant D(x,y)=fxxfyy−(fxy)2.
D(x,y)=(2)(2)−(1)2=4−1=3
Step 4: Apply the Second Derivative Test to the critical point.
- At (−34,−31):
D(−34,−31)=3
Since D>0, we check fxx:
fxx(−34,−31)=2
Since fxx>0, the critical point (−34,−31) is a local minimum.
Step 5: Calculate the function value at the local minimum.
f(−34,−31)=(−34)2+(−34)(−31)+(−31)2+3(−34)+2(−31)+5
f(−34,−31)=916+94+91−312−32+5
f(−34,−31)=921−314+5
f(−34,−31)=37−314+315
f(−34,−31)=37−14+15=38
The final answer is Localminimumat(−34,−31)withvalue38.
3 done, 2 left today. You're making progress.