Question

find the maximum value of the function f(x,y,z)=x+2y+3z on the curve of interjection of the plane x-y+z=1 and cylinder x2+y2=1

Answer 1

find the maximum value of the function f(x,y,z)=x+2y+3z on the curve of interjection of the plane x-y+z=1 and cylinder x2+y2=1

Answer 2

i recall that the normal of a surface always points in the steepest direction

Answer 3

well, the gradF that is ...

Answer 4

This is a constrained maximization problem. You are looking for the maximum of f constrained to the the two equation that you have given. I think the best way is to use Lagrange multipliers. Is this calculus 2? I can tell you more later if you want.

Answer 5

another approach is to recognize that x^2 +y^2 =1 is the unit circle
the critical points of "f" are along outermost points of constrained region
this represents a slanted circle where x,y are points along unit circle and z is defined by the given plane (z = 1+y-x)

make substitution:
\[x = \cos \theta\]
\[y = \sin \theta\]
\[z = 1+\sin \theta - \cos \theta\]
now you can define "f" in terms of just angle "theta" and find angle which maximizes "f"
\[f(\theta) = 5\sin \theta -2\cos \theta +3\]
\[f'(\theta) = 5\cos \theta +2\sin \theta = 0\]
\[\rightarrow \tan \theta = -\frac{5}{2}\]
\[\theta_1 \approx 111.8\]
\[\theta_2 \approx 291.8\]
one will yield max the other min
\[f(\theta_1) = 8.385\]
\[f(\theta_2) = -2.385\]
therefore maximum is 8.385 with angle of 111.8 degrees