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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)
thanks for your big help....!
Quote:
Originally Posted by ggw
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)
I hope the following attachment helps, because this is only half of the solution.
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=0Quote:
Originally Posted by ggw
f_y=-6x+2y=0
(x,y)={(0,0),(6,18)}
The only one inside the triangle is,
(0,0).
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.
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...!