# find absolute extrema & Lagrange Multipliers

• Mar 30th 2007, 10:05 PM
ggw
find absolute extrema & Lagrange Multipliers
1. given: f(x,y) = x^3 - 6xy + y^2 +15x

find absolute extrema of f over the triangluar region in the xy plane with vertices (0,0), (0,4) & (2,4).

2. given: T(x,y,z)=100+x^2 + y^2 represent the temperature at each point on the sphere x^2 + y^2 + z^2 =50. Use Lagrange Multipliers to find the Maximum & Minimum temperature on the curve formed by the intersection of the sphere and the plane x - z = 0. (hint: Optimizing a function subject to two constraints)

• Mar 31st 2007, 05:07 PM
ThePerfectHacker
Quote:

Originally Posted by ggw
1. given: f(x,y) = x^3 - 6xy - y^2 +15x

find absolute extrema of f over the triangluar region in the xy plane with vertices (0,0), (0,4) & (2,4).

First let us find the equation of each side of the triangle.

(0,0),(0,4)
It is the equation of the line in the xy plane which connects those two.
Hence,
y=4.

(0,0),(2,4)
y=2x

(0,0),(0,4)
x=0

Inside Triangle
f_x=0
f_y=0

Thus,

3x^2-6y+15=0
-6x-2y=0

Solve, (you are big enough to solve the system thyself).

There exists solutions but not inside the open boundary.

Hence we check the boundary itself.

Line y=4
f(x,4)=x^3-24x-16+15x=x^3-9x-16
Find extreme values on (0,2) and you find there are none.
Now check its boundaries x=0 and x=2
Hence,
f(0,4)=-16 and f(2,4)=8-18-16=-26

Line x=0
f(0,y)=-y^2
Find extreme values on (0,4) and there are none.
Now check its boundaries y=0 and y=4.
f(0,0)=0 and f(0,4)=-16

Line y=2x
f(x,y)=x^3-12x^2-4x^2+15x=x^3-16x^2+15x
Find extreme values on (0,2) and there are none.
Now check its boundaries x=0 and x=2.
f(0,0)=0 and f(2,4)=8-64+30=-26

The smallest values was at,
(2,4)
The largest value was at,
(0,0)
• Mar 31st 2007, 05:32 PM
ThePerfectHacker
I hope the following attachment helps, because this is only half of the solution.
• Apr 1st 2007, 06:55 PM
ggw
missing plus sign
for the f(x,y) = x^3 -6xy + y^2 +15

the function should be like this, and the critical point should be x =1, y=3. I dont know are there any other criticals point might have in the function.
• Apr 1st 2007, 07:02 PM
ThePerfectHacker
Quote:

Originally Posted by ggw
for the f(x,y) = x^3 -6xy + y^2 +15

the function should be like this, and the critical point should be x =1, y=3. I dont know are there any other criticals point might have in the function.

f_x=3x^2-6y=0
f_y=-6x+2y=0

(x,y)={(0,0),(6,18)}
The only one inside the triangle is,
(0,0).
• Apr 2nd 2007, 06:36 PM
ggw
here is what i did
Partial Derivatives
fx =3x^2 -6y + 15
fy =-6x+2y solve for x & y i got. x = 1, y =3
=> critical points is (1,3)
f(1,3) = 7
Line 1: y =4 0<=x<=2
f(x,4)=x^3 -9x +16
so, min is f(0,0) = 0, max is f(2,0)=6
Line 2: x = 0 0<=y<=4
f(0,y) = y^2
so, min is f(0,0) = 0, max is f(0,4) =16
Line 3: y = 2x 0<=x<=2
f(x,y) = x^3 -8x^2 + 5x
so, min is f(0,0) =0, f(2,0) = -14 there are none

what would i conclude this problem, and am I corretct ?

The second questions, I am not sure if i understand well. Thanks for your help.
• Apr 3rd 2007, 12:35 PM
ggw
i am still woring out on problem 2,

here is what i got

y = +- sqrt(50)

and (0,150,0) is relative max. Not really sure about it, need help...!